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Modern Physics for Scientists and Engineers
Solutions
Joseph N. Burchett after consultations with John C. Morrison
August 4, 2010
Copyright © 2010 by Elsevier, Inc. All rights reserved.
2
Copyright © 2010 by Elsevier, Inc. All rights reserved.
Introduction - Solutions
1. We are given a spherical distribution of charge and asked to calculate the
potential energy of a point charge due to this distribution. There are two
situations to consider which deliver quite diﬀerent results. The problem does
not specify whether the point charge of interest is inside or outside the charge
distribution so we must take both possibilities into consideration. Before we
get started, let’s set up an appropriate description of our charge distribution
so that we may proceed with each of the above situations. It is most useful
to express the charge distribution in terms of a charge density ρ:
ρ =
Q
4
3
πR
3
This is the charge per unit volume as found by merely dividing the total
charge Q by the total volume of the sphere. We now handle the “outside the
sphere” conﬁguration. All of the charge inside the sphere may be considered
as though it is concentrated at a single point. We may then use the Coulomb’s
Law result for the force acting on the point charge q:
F =
1
4π
0
Qq
r
2
We integrate to ﬁnd the potential energy:
V = −
Qq
4π
0
_
r
∞
1
r
2
dx =
Qq
4π
0
1
r
This result is also identical to the one obtained in the text in the case of the
two-point charge conﬁguration. We see that this should in fact be the case
since we started with the same quantity of force.
Now, to handle the “inside the sphere” situation, we shall use the above
stated charge density. We easily see that the distance r from the center of the
3
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4
distribution of charge to the point charge is less than the radius R. Imagine
now a sphere of radius r

that is inside the larger spherical charge distribution
but centered at the same point. Only the charge within this inner sphere will
act on the point charge. We may now use our charge density to ﬁnd the
charge contained in the smaller sphere:
Q
in
=
4
3
πr
3
ρ
The force on a charge q due to the inner sphere is then given by Coulomb’s
Law:
F =
q
4π
0
4πr
3
ρ
3r
2
=
qρr

3
0
Before we integrate to ﬁnd the potential energy, we must make an important
distinction from the “outside the sphere” procedure above. We were able to
integrate directly from inﬁnity to the location of our point charge because the
amount of source charge taken into account did not change (Q). However, as
we enter the sphere, the amount of source charge begins to decrease until we
reach our point at distance r from the center. We split the integration into
two parts corresponding to both regions to ﬁnd the potential energy:
V = −[
_
R
∞
Qq
4π
0
1
r
2
dr

+
_
r
R
qρr

3
0
dr

]
=
Qq
4π
0
1
R
−
ρq
6
0
(r
2
−R
2
)
Note that the ﬁrst integral above was identical to the one for the outside
conﬁguration, just evaluated at r = R. Now, we just substitute our charge
density ρ and simplify:
V =
Qq
4π
0
1
R
−
Qqr
2
8π
0
R
3
+
Qq
8π
0
R
=
3Qq
8π
0
R
−
Qqr
2
8π
0
R
3
3. Looking at the deﬁnition of kinetic energy:
KE =
1
2
mv
2
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5
Let’s solve for velocity:
v =
_
2(KE)
m
Let KE

= 4KE.
v =
_
2(KE

)
m
=
_
2(4KE)
m
= 2
_
2(KE)
m
= 2v
So, the speed will increase by a factor of 2 if the kinetic energy is increased
by a factor of 4. By the deﬁnition of momentum:
p = mv ⇒m(2v) = 2p
We should also expect the momentum of the particle above to increase by
a factor of 2. In fact, as long as the mass does not change, the momentum
should increase or decrease by the same factor as the changing speed.
5. The frequency and wavelength of light are related by equation I.23:
f =
c
λ
The constant c refers to the speed of light which we know to be 3.00×10
8
m/s.
Taking care to convert our wavelength, which is given in nanometers, to
meters:
f =
3.00 ×10
8 m
s
500 ×10
−9
m
= 6 ×10
14
Hz
7. Equation I.25 relates the photon energy to the wavelength:
E =
hc
λ
Substituting the value hc = 1240 eV ·nm:
E =
1240 eV ·nm
500nm
= 2.48 eV
9. The energy of a quantum of light (the photon) must be equal to the
diﬀerence in energies of the two states:
E
photon
= E
2
−E
1
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6
According to equation I-25, the energy also obeys the following:
E
photon
=
hc
λ
Equating the two and solving for λ, we get:
λ =
hc
E
2
−E
1
Copyright © 2010 by Elsevier, Inc. All rights reserved.
1
The Wave-Particle Duality -
Solutions
1. The energy of photons in terms of the wavelength of light is given by eq.
1.5. Substituting the given wavelength into eq. 1.5:
E
photon
=
hc
λ
=
1240 eV · nm
200nm
= 6.2 eV
3. Considering that we are given the power of the laser in units of milli-
watts, the power maybe expressed as: 1 mW = 1 × 10
−3
J/s. This gives us
a strong hint how to proceed. We are hoping to ﬁnd the rate of emission
of photons, so that if we can ﬁnd the energy of a single photon, we can use
the power as stated above to calculate the number of photons. The energy
of a single photon can be calculated as in Problem 1 by substituting the
wavelength of the light into equation 1.5:
E
photon
=
hc
λ
=
1240 eV · nm
632.8 nm
= 1.960 eV
We now convert to SI units:
1.960 eV ×
1 J
1.6 ×10
−19
eV
= 3.14 ×10
−19
J
Now, using the given laser power:
Rate of emission =
power
energy per photon
=
1 ×10
−3 J
s·photon
3.14 ×10
−19
J
= 3.18×10
15
photons
s
7
Copyright © 2010 by Elsevier, Inc. All rights reserved.
8 1. THE WAVE-PARTICLE DUALITY - SOLUTIONS
5. This problem concerns the photoelectric eﬀect. Given the work function
of the material and the emitted electron kinetic energy, we wish to calculate
the wavelength of the light incident to the surface. Equation 1.6 provides
the following:
(KE)
max
=
hc
λ
−W where W is the work function of the material.
The hc/λ term describes the energy supplied by the incoming photons. By
viewing the work function as an energy threshold for producing photoelectric
current, we see that the amount by which the photon energy (hc/λ) exceeds
the work function is the resultant maximum kinetic energy of the emitted
electrons. We may thus write:
hc
λ
= (KE)
max
+W
= 2.3 eV + 0.9 eV
= 3.2 eV
Using hc = 1240 eV · nm,
λ =
1240 eV · nm
3.2 eV
= 387.5 nm
6. Here, we are given the stopping potential of a photoelectric experiment
and wish to determine the work function of the metal. Since 0.72 eV is the
necessary potential energy to cease the ﬂow of electrons, the maximum ki-
netic energy of the electrons being emitted must equal 0.72 eV. Our problem
is then reduced to solving eq. 1.6 for the work function:
W =
hc
λ
−(KE)
max
=
1240 eV · nm
460 nm
−0.72 eV = 1.98 eV
11. For this atomic transition, the energy of the emitted photon must equal
Copyright © 2010 by Elsevier, Inc. All rights reserved.
9
the diﬀerence in energy of the two states of hydrogen (n = 2 and n = 5).
Equation 1.22 gives us those energies, thus:
E
photon
= E
5
−E
2
= −
13.6 eV
5
2
−(−
13.6 eV
2
2
) = 2.86 eV
From eq. 1.12:
λ =
hc
∆E
=
1240 eV · nm
2.86 eV
= 434.2 nm
15. We must ﬁrst ﬁnd the energy of the UV photons, which may be done
via eq. 1.5:
E
photon
=
hc
λ
=
1240 eV · nm
45 nm
= 27.6 eV
If electrons are to be emitted, we need to overcome the energy that is keeping
them in a bound state. For hydrogen atoms, equation 1.22 gives us that
energy. We need only supply the n quantum number for the state of the
atom (for ground state, n = 1):
E = −
13.6 eV
n
2
= −
13.6 eV
1
2
= −13.6 eV
The kinetic energy of these emitted electrons should then be equal to the
diﬀerence between the energy provided by the incident light and the ground
state electron energy: KE = 27.6 eV − 13.6 eV = 14.0 eV For the velocity
calculation, it will be useful to convert from eV to J:
14.0 eV ·
1.6 ×10
−19
J
1 eV
= 2.24 ×10
−18
J (1.1)
We can then ﬁnd the velocity using the deﬁnition of kinetic energy and
the mass of the electron:
KE =
1
2
mV
2
⇒v =
_
2(KE)
m
=
¸
2.24 ×10
−18
J
9.11 ×10
−31
kg
= 2.21 ×10
6
m
s
(1.2)
17. It should be ﬁrst noted that the wavelength of the emitted light increases
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10 1. THE WAVE-PARTICLE DUALITY - SOLUTIONS
as the photon energy decreases. Thus, the longest acceptable wavelength
will correspond to a transition to the nearest excited state. Transitions to
higher-lying states will require more energy and thus correspond to a shorter
wavelength. We then examine the energy of the transition from the n = 3
state of the hydrogen atom. Equation 1.22 gives:
E
3
= −
13.6 eV
n
2
= −
13.6 eV
3
2
= −1.51 eV
The next available state is the state corresponding to n=4 whose energy may
be found by eq. 1.22 again:
E
4
= −
13.6 eV
n
2
= −
13.6 eV
4
2
= −0.85 eV
So, in order for the absorption to take place, the incident photons must
contain at least as much energy as the diﬀerence in these energies (−0.85 eV −
(−1.51 eV ) = .66 eV ). We may now solve eq. 1.5 for the wavelength:
λ =
hc
∆E
=
1240 eV · nm
.66 eV
= 1878 nm
Thus, any light of wavelength greater than 1878 nm would not possess suﬃ-
cient energy to be absorbed by the hydrogen atom.
19. Using a procedure such as that of Example 1.5, we can relate the kinetic
energy to the de Broglie wavelength by the following equation:
λ =
h
_
2m(KE)
We need only convert the kinetic energy from eV to J and substitute the
appropriate masses.
KE = 20 eV ×
1.6 ×10
−19
J
1 eV
= 3.2 ×10
−19
J
Electron : λ =
6.626 ×10
−34
J·s
_
2(9.11 ×10
−31
kg)(3.2 ×10
−18
J)
= 2.74 ×10
−10
m
Proton : λ =
6.626 ×10
−34
J·s
_
2(1.67 ×10
−27
kg)(3.2 ×10
−18
J)
= 6.41 ×10
−12
m
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11
An α-particle is the most commonly occurring isotope of the helium atom
consisting of two protons and two neutrons. We can thus approximate its
mass: m = 2(1.672 ×10
−27
) + 2(1.674 ×10
−27
) = 6.692 ×10
−27
kg.
α −particle : λ =
6.626 ×10
−34
J·s
_
2(6.692 ×10
−27
kg)(3.2 ×10
−18
J)
= 3.20 ×10
−12
m
24. The Davisson-Germer experiment measured the scattering of electrons
by a crystal during which the interference patterns characteristic of light were
shown to also occur with particles. We can solve Bragg’s Law (eq. 1.24) for
the inter-atom spacing of the crystal in terms of the de Broglie wavelength
and the scattering angle:
2d sin(θ) = nλ ⇒d =
nλ
2 sin(θ)
We may use the relation derived in example 1.5 to ﬁnd the de Broglie wave-
length from kinetic energy after ﬁrst converting our kinetic energy to SI
units:
KE = 54 eV ×
1.6 ×10
−19
1 eV
λ =
h
_
2m(KE)
=
6.626 ×10
−10
m
_
2(9.11 ×10
−31
kg)(8.64 ×10
−18
J)
= 1.66 ×10
−10
m
We let n=1 and solve Bragg’s Law for d, the spacing in the crystal:
d =
λ
2 sin(θ)
=
1.66 ×10
−10
m
2 sin(50
◦
)
= 1.09 ×10
−10
m
Copyright © 2010 by Elsevier, Inc. All rights reserved.
12 1. THE WAVE-PARTICLE DUALITY - SOLUTIONS
Copyright © 2010 by Elsevier, Inc. All rights reserved.
2
The Schr¨ odinger Wave
Equation - Solutions
1. Equation 2.17 gives the energies for the particles in an inﬁnite potential
well:
E =
n
2
h
2
2mL
2
Part (a) asks for the size of the region, which we can ﬁnd by solving eq. 2.17
for L:
L =
_
n
2
h
2
8mE
We need only supply some information such as the energy (1.0 eV as given),
the mass of the electron (9.11 ×10
−31
kg from Appendix A), and the appro-
priate value of n. The lowest energy will be found when the electron is at
the ground state (n=1). We must also convert 1.0 ev to 1.6 ×10
−19
J. Now
we are ready:
L =
¸
1
2
(6.63 ×10
−34
J·s)
2
8(9.11 ×10
−31
kg)(1.6 ×10
−19
J)
= 6.14 ×10
−10
m
While the derivation of eq. 2.17 is correct, let’s do a little dimensional anal-
ysis to better “see through” our length calculation above. The quantum
number n is dimensionless as is of course the constant 8, so let’s examine the
operations of units of the rest of the quantities:
_
n
2
h
2
8me
⇒
¸
J
2
· s
2
kg · J
=
¸
J · s
2
kg
13
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14 2. THE SCHR
¨
ODINGER WAVE EQUATION - SOLUTIONS
Now substitute 1 J = 1 kg
m
2
s
2
· s
2
:
¸
J · s
2
kg
=
¸
kg
m
2
s
2
· s
2
kg
=
√
m
2
= m
It should be emphasized that the m here is the abbreviation for the unit of
length meters, not the quantity of mass.
Part (b) can be approached in a couple of diﬀerent ways. The essential
quantity we must obtain now is the energy of the ﬁrst level above ground
state (also called the ﬁrst excited state). The energy required to excite the
electron must be:
∆E = E
n2
−E
n1
= E
2
−E
1
The quantum numbers n=2 and n=1 correspond to the ﬁrst excited state
and ground state respectively. So, for E
2
from equation 2.17:
E
2
=
2
2
h
2
8mL
2
=
(6.63 ×10
−34
J·s)
2
2(9.11 ×10
−31
kg)(6.14 ×10
−10
m)
2
= 2.14 ×10
−18
J ×
1 eV
1.6 ×10
−19
J
= 4.00 eV
Here we used the result of the size of the well from part (a). The other
method would involve solving eq. 2.17 for everything except n and using our
given quantities as follows:
E =
n
2
h
2
2mL
2
⇒
h
2
8mL
2
=
E
n
2
h
2
8mL
2
= (1.0 eV )(1
2
) = 1.0 eV
E
n
= n
2
(1.0 eV ) ⇒E
2
= 4.0 eV
Now for our desired result:
∆E = 4.0eV −1.0eV = 3.0eV
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15
3. As with the atomic transitions we dealt with in chapter 1, the emit-
ted photon energy will equal the diﬀerence of the energies of the two states
corresponding to n = 3 and n = 2. So our task becomes ﬁnding each of these
energies via equation 2.17:
E
3
=
3
2
h
2
8mL
2
=
9(6.63 ×10
−34
J·s)
2
8(9.11 ×10
−31
kg)(10 ×10
−9
m)
2
= 5.43 ×10
−21
J ·
1 eV
1.6 ×10
−19
J
= .034 eV
E
2
=
2
2
h
2
8mL
2
=
4(6.63 ×10
−34
J·s)
2
8(9.11 ×10
−31
kg)(10 ×10
−9
m)
2
= 2.41 ×10
−21
J ·
1 eV
1.6 ×10
−19
J
= .015 eV
∆E = .034 eV −.015 eV = .019 eV
This value of ∆E will then be equal to the energy of the photon and we may
use eq. 1.5 to calculate the wavelength of the light:
KE =
hc
λ
⇒λ =
hc
KE
λ =
1240 eV · nm
.019 eV
= 65263 nm
5. Note the typo, “Draw the wave function...” should be “Find the wave
function...”
We are given the following information: ﬁnite well of depth 0.3 eV and
thickness L = 10 nm, we are dealing with conduction electrons if GaAS
in their lowest state, and these electrons have an eﬀective mass of 0.067
times the electron mass. From section 2.3, we see the process of solving
the ﬁnite square well which culminates in the numerical solution of tanθ =
_
(θ
2
0
/θ
2
) −1, for the even solutions, and −cotθ =
_
(θ
2
0
/θ
2
) −1 for the odd
case. Our variables θ and θ
0
correspond to the following expressions:
θ =
kL
2
, θ
0
2
=
mV
0
L
2
2
2
(from eq. 2.30)
Notice that the latter of the two is a squared quantity where the former is
not, but we’ll deal with that a little later on. We are interested in the ground
Copyright © 2010 by Elsevier, Inc. All rights reserved.
16 2. THE SCHR
¨
ODINGER WAVE EQUATION - SOLUTIONS
state of the electron, which will correspond to n=1, leading us to an even
solution which may be found by equation 2.29:
tanθ =
¸
θ
0
2
θ
2
−1
In the text near the end of section 2.3, the value of θ
0
2
is given to be 13.2.
Our equation becomes:
tanθ =
_
13.2
2
θ
2
−1
By using a computer algebra system or calculator to graph the left-hand and
right-hand sides, we ﬁnd the ﬁrst intersection at θ ≈ 1.22. Thus, we may
solve θ = kL/2 for k:
k =
2θ
L
=
2(1.22)
10 ×10
−9
m
= 2.44 ×108 m
−1
For the equations outside the well, we must solve for κ as well. Let’s ﬁrst
solve equation 2.24 for E:
k =
2mE

2
2
⇒E =
k
2

2
2m
=
(2.44 ×10
8
m
−1
)
2
(1.054 ×10
−34
J · s)
2
2(6.103 ×10
−32
)
kg
E = 5.419/times10
−21
J ×
1 eV
1.6 ×10
−19
J
= .034 eV
Now for κ (from eq. 2.26):
κ =
¸
2(6.103 ×10
32
kg)(.3 eV −.034 eV )(1.6 ×10
−19
J
eV
)
(1.054 ×10
−34
J · s)
2
= 6.84×10
−8
m
−1
We know have our necessary information for the wave equations:
Ψ(x) =
_
_
_
B exp(6.84 ×10
−8
m
−1
)x : x ≤ −
L
2
A cos(2.44 ×10
−8
m
−1
)x : −
L
2
≤ x ≤
L
2
B exp(6.84 ×10
−8
m
−1
)x : x ≥
L
2
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17
9. Outside the ﬁnite potential well, V = V
0
, thus our time-independent
Schrodinger equation is:
−

2
2m
d
2
ψ
dx
2
+V
0
ψ = Eψ
If we multiply both sides by −2m/
2
:
d
2
ψ
dx
2
−
2mV
0

2
ψ = −
2mE

2
ψ
Moving 2mE/
2
to the LHS:
d
2
ψ
dx
2
+
2m(E −V
0
)

2
ψ = 0
We make the following substitution:
k =
_
2m(E −V
0
)

2
_1
2
Since E is greater than V
0
, k is in fact a real number and our Schrodinger
equation becomes:
d
2
ψ
dx
2
+k
2
ψ = 0
And, we may readily conﬁrm by substitution that the general form of the
solution of this equation is a linear combination of the functions, A cos(kx)
and B sin(kx), and is oscillatory in nature.
11. The normalized wave function must satisfy the normalization condition
(equation 2.18):
_
∞
−∞
|ψ(x)|
2
dx = 1
_
∞
−∞
Ae
−mωx
2
\2
Ae
−mωx
2
\2
dx =
_
∞
−∞
A
2
e
−mωx
2
\
Therefore,
1
A
2
=
_
∞
−∞
e
−mωx
2
\
dx = 2
_
∞
0
e
−mωx
2
\
dx
Copyright © 2010 by Elsevier, Inc. All rights reserved.
18 2. THE SCHR
¨
ODINGER WAVE EQUATION - SOLUTIONS
This looks very similar to the integral we are given if we let a = mω/. Then:
1
A
2
=
_
π
mω

=
_
π
mω
A =
_
mω
π
_1
4
13. a) From 0 ≤ x ≤ L, V = 0, and our Schrodinger equation becomes:
−

2
2m
d
2
ψ
dx
2
= Eψ or
d
2
ψ
dx
2
+
2mE

2
ψ = 0
For x ≥ L, V = V
0
and:
−

2
2m
d
2
ψ
dx
2
+V
0
= Eψ or
d
2
ψ
dx
2
−
2(V
0
−E)m

2
ψ = 0
b) If we let:
k =
_
2mE

2
and κ =
_
2m(V
0
−E)

2
Then,
d
2
ψ
dx
2
+k
2
ψ = 0 ; 0 ≤ x ≤ L
d
2
ψ
dx
2
−k
2
ψ = 0 ; x ≥ L
Similar to the situation of the ﬁnite well discussed in the text, the above
equations are thus satisﬁed by the following:
ψ(x) = A cos(kx) and ψ(x) = A sin(kx) ; 0 ≤ x ≤ L
ψ(x) = Be
−κx
; x ≥ L
Once again, the negative argument in the exponential is to insure our wave
function decays as x →∞.
c) If the potential is to be inﬁnite at x = 0, then the wave function must
go to zero at x = 0. Therefore, we must impose new boundary conditions
to insure the continuity of the wave function. Above we had both even and
odd solutions for the wave function, so let’s examine their behavior at x = 0
given the new conditions:
Even : ψ(0) = A cos(k0) = 0
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19
Odd : ψ(0) = A sin(k0) = 0
We see immediately that the ﬁst of the equations is not physically acceptable,
so we must reject the even solutions, thus we are left with: ψ(x) = A sin(kx)
for 0 ≤ x ≤ L. We may now proceed with our boundary conditions at x = L:
ψ(L) = A sin(kL) = Be
−κL
Imposing the continuity of the ﬁrst derivative:
Ak cos(kL) = −Bκe
−κL
Divide eq. 2 by eq. 1:
k cot(kL) = −κ
−cot(kL) =
κ
k
We insert our κ as deﬁned in part (b):
−cot(kL) =
_
2mV
0

2
k
2
−
2mE

2
k
2
Substitute k into the second term of the RHS:
−cot(kL) =
_
2mV
0

2
k
2
−1
Let θ = kL:
−cot(θ) =
_
2mV
0
L
2

2
θ
2
−1
We can transform this into a form similar to that of eq. 2.31 if:
θ
0
2
=
2mV
0
L
2

2
Our equation to solve then becomes:
−cot(θ) =
¸
θ
0
2
θ
2
−1
This can be solved graphically for the values of θ for which the LHS and RHS
expressions intersect. From here we obtain the corresponding values of k:
k =
θ
L
Copyright © 2010 by Elsevier, Inc. All rights reserved.
20 2. THE SCHR
¨
ODINGER WAVE EQUATION - SOLUTIONS
And solve eq. 2.24 for E:
E =
k
2

2
2m
Note that the value for θ
0
2
which we derived above is slightly diﬀerent from
that of eq. 2.30. This occurs because our potential well for this problem is
actually oriented from x = 0 to x = L, as opposed to x = −L/2 to x = L/2
in the example of section 2.3. This in turn gave us a diﬀerent argument of the
cotangent function and thus necessitated we choose a diﬀerent θ
0
2
in order
to keep the same clean form of eq. 2.31 to solve numerically. Procedurally,
the process we just followed is identical to that of section 2.3.
15. The inﬁnite potential well represents a situation where the potential
energy does not evolve with time. The time-dependent solutions must then
satisfy eq. 2.45:
Ψ(x, t) = u
E
(x)e
−iωt
where w =
E

From eq. 2.20, the normalized spatial wave functions of a particle in the
inﬁnte well are given by:
ψ(x) =
_
2
L
sin
_
nπx
L
_
The corresponding energies are given by eq. 2.17:
E =
n
2
h
2
8mL
2
We may then calculate ω from eqs. 2.39 and 2.17:
ω =
E

=
n
2
πh
4mL
2
Our total wave function is then:
Ψ(x, t) =
_
2
L
sin
_
nπx
L
_
exp
_
−
in
2
πht
4mL
2
_
Copyright © 2010 by Elsevier, Inc. All rights reserved.
3
Operators and Waves -
Solutions
1. We may obtain the momentum by multiplying the momentum operator
as given in equation 3.2
ˆ p = −i
d
dx
times the wave function given in the problem
ˆ pψ(x) = −i
d
dx
Ae
iαx
= −i
2
αAe
iαx
ˆ pψ(x) = αAe
iαx
Our wavefunction is therefore seen to be an eigenfunction of the momentum
operator corresponding to the eigenvalue α. Thus, a measurement of the
momentum of the electron in this state would yield a value of α.
The kinetic energy may be found via the relation:
KE =
p
2
2m
The square of the momentum operator that appears here in the numerator
will result in the operator begin applied twice, therefore:
ˆ p
2
= (−i)
2
d
2
dx
2
= −
2
d
2
dx
2
We substitute this expression into the expression of the kinetic energy:
KE =
ˆ p
2
2m
= −

2
2m
d
2
dx
2
21
Copyright © 2010 by Elsevier, Inc. All rights reserved.
22 3. OPERATORS AND WAVES - SOLUTIONS
and operate on the wavefunction:
−

2
2m
d
2
ψ
dx
2
= −

2
2m
d
2
dx
2
Ae
iαx
=

2
2m
Aα
2
e
iαx
⇒KE =

2
α
2
2m
3. Our task is to ﬁnd ψ = Aφ
1
(x) +Bφ
2
(x) such that:
ˆ pψ = pψ
Where p is the eigenvalue of the momentum operator. So, let’s try operating
on the following wavefunction:
ψ = A cos(kx) +B sin(kx)
ˆ pψ = −i
dψ
dx
= −iA
d
dx
cos(kx) −iB
d
dx
sin(kx)
= ik Asin(kx) −ik cos(kx)
This choice of ψ is not generally an eigenfunction of ˆ p, as we can not factor
out our original ψ = A cos(kx) +B sin(kx). Let us now try the following:
ψ = cos(kx) +i sin(kx)
ˆ pψ = −ik(−sin(kx)) −i
2
k(cos(kx)) = k cos(kx) +ik sin(kx)
= k(cos(kx) +i sin(kx))
Thus, we have found an eigenfunction of the momentum operator which
corresponds to the eigenvalue k. Note, that using the Euler formula:
e
ikx
= cos(kx) +i sin(kx)
Our wavefunction may be written:
ψ = e
ikx
Using this form of the wavefunction and the result of Problem 1, we may
Copyright © 2010 by Elsevier, Inc. All rights reserved.
23
thus see that the wavefunction is an eigenfunction of the momentum opera-
tor corresponding to the eigenvalue k.
5. Notice that if the energy is less than the minimum value of V (x), the
[V (x) −E] portion of the RHS is always positive. Therefore, if this equation
holds, the second derivative of the wavefunction will have the same sign as
the wavefunction itself, and the function will increase or decrease in value
monotonically.
7. Sections 3.3.1 and 3.3.2 provide the initial setup for this problem. We
must solve the appropriate Schrodinger equation for each region, then use
boundary conditions to ﬁnd our desired ratios. In the ﬁrst region, where
V = 0, we have the following from section 3.3.1:
d
2
ψ
1
dx
2
+k
1
2
ψ
1
= 0
and given by eq. 3.32:
k
1
=
_
2mE

2
This yields solutions:
ψ
1
= Ae
ik
1
x
+Be
−ik
1
x
Region 2 produces equations 3.39 through 3.40:
d
2
ψ
2
dx
2
−k
2
2
ψ
2
= 0 where k
2
=
_
2m(V
0
−E)

2
With physically acceptable solutions:
ψ
2
(x) = De
−k
2
x
Now, we impose continuity of the wave equations at x = 0 and of their ﬁrst
derivatives at x = 0:
ψ
1
(0) = ψ
2
(0) ⇒A +B = D
ψ
1

(0) = ψ
2

(0) ⇒ik
1
(A −B) = −k
2
D
Divide both side of the second equation by ik
1
and our two equations become:
A +B = D (1)
Copyright © 2010 by Elsevier, Inc. All rights reserved.
24 3. OPERATORS AND WAVES - SOLUTIONS
A −B = −
k
2
ik
1
D = i
k
2
k
1
D (2)
Here, the fact that −1/i = i was used to bring i out of the denominator and
cancel the negative. If we add these two equations, we solve for A in terms
of D:
2A = D(1 +i
k
2
k
1
)
A =
1
2
D(1 +i
k
2
k
1
) (3)
If we subtract (1) from (2), we can solve for B in terms of D:
2B = D(1 −i
k
2
k
1
)
B =
1
2
D(1 −i
k
2
k
1
) (4)
From here, we divide each (3) and (4) by D:
A
D
=
1
2
(1 +i
k
2
k
1
)
B
D
=
1
2
(1 −i
k
2
k
1
)
The reﬂection coeﬃcient R is found by
R =
|B|
2
v
1
|A|
2
v
1
=
|B|
2
|A|
2
Using our ratios above:
R =
(
B
D
)
∗
(
B
D
)
(
A
D
)
∗
(
A
D
)
=
1
4
(1 −i
k
2
k
1
)(1 +i
k
2
k
1
)
1
4
(1 +i
k
2
k
1
)(1 −i
k
2
k
1
)
= 1
9. Following ﬁgure 3.6, we are given the general solutions for each region:
ψ
1
(x) = Ae
ik
1
x
+Be
−ik
1
x
, x ≤ 0
ψ
2
(x) = Ce
k
2
x
+De
−k
2
x
, 0 ≤ x ≤ L
ψ
3
(x) = Ee
ik
1
x
, x ≥ 0
Copyright © 2010 by Elsevier, Inc. All rights reserved.
25
We impose the boundary conditions at x = 0 and x = L:
ψ
1
(0) = ψ
2
(0)
ψ
1

(0) = ψ
2

(0)
ψ
2
(L) = ψ
3
(L)
ψ
2

(L) = ψ
3

(L)
The ﬁrst two yield the following system of equations:
A +B = C +D
ik
1
(A −B) = k
2
(C −D)
We divide the second of these by ik
1
and add the result to the ﬁrst to obtain
an expression for A in terms of C and D:
2A = C +D −i
k
2
k
1
(C −D) (1)
Now, let’s turn to the interface between regions 2 and 3. The boundary
conditions produce the following system:
Ce
k
2
L
+De
−k
2
L
= Ee
k
1
L
k
2
Ce
k
2
L
−k
2
De
−k
2
L
= ik
1
Ee
k
1
L
Dividing the second of these equations by k
2
reduces to the system to a more
manageable one:
Ce
k
2
L
+De
−k
2
L
= Ee
k
1
L
Ce
k
2
L
+De
−k
2
L
= i
k
1
k
2
Ee
k
1
L
Add the two equations and we can solve for C in terms of E:
2Ce
k
2
L
= Ee
k
1
L
+i
k
1
k
2
Ee
k
1
L
C =
E
2
_
1 +i
k
1
k
2
_
e
(ik
1
−k
2
)L
(2)
Copyright © 2010 by Elsevier, Inc. All rights reserved.
26 3. OPERATORS AND WAVES - SOLUTIONS
Subtract those same equations to solve for D in terms of E:
2De
−k
2
L
= Ee
k
1
L
−i
k
1
k
2
Ee
k
1
L
D =
E
2
_
1 −i
k
1
k
2
_
e
(ik
1
+k
2
)L
(3)
Insert (2) and (3) into (1):
2A =
E
2
_
1 +i
k
1
k
2
_
e
(ik
1
−k
2
)L
+
E
2
_
1 −i
k
1
k
2
_
e
(ik
1
+k
2
)L
−i
k
2
k
1
E
2
__
1 +i
k
1
k
2
_
e
(ik
1
−k
2
)L
+
_
1 −i
k
1
k
2
_
e
(ik
1
+k
2
)L
_
Divide both sides by A and simplify:
4 =
E
A
__
1 +i
k
1
k
2
_
−
_
i
k
1
k
2
−1
__
e
ik
1
L−k
2
L
+
E
A
__
1 −i
k
1
k
2
_
+
_
i
k
1
k
2
+ 1
__
e
ik
1
L+k
2
L
=
E
A
__
2 +i
k
2
1
+k
2
2
k
1
k
2
_
e
ik
1
L−k
2
L
+
_
2 +i
k
2
2
−k
2
1
k
1
k
2
_
e
ik
1
L+k
2
L
_
Solve for E/A:
E
A
= 4
__
2 +i
k
2
1
+k
2
2
k
1
k
2
_
e
ik
1
L−k
2
L
+
_
2 +i
k
2
2
−k
2
1
k
1
k
2
_
e
ik
1
L+k
2
L
_
−1
11. The Heisenberg uncertainty principle states:
∆E∆t ≥

2
Therefore we solve for ∆E:
∆E =

2∆t
=
6.626 ×10−34 J · s
2(4.0 ×10
−10
s)
= 8.28 ×10
−25
J
Copyright © 2010 by Elsevier, Inc. All rights reserved.
27
Now, converting to electron volts:
∆E = 8.28 ×10
−25
J ·
1 eV
1.6 ×10
−19
J
= 5.18 ×10
−6
eV
13. Equation 3.48 gives the average value of an observable:
< Q >=
_
ψ
∗
(x)
ˆ
Qψ(x) dx
Where
ˆ
Q is the corresponding operator, in this case the kinetic energy oper-
ator, which is given by:
ˆ p
2
2m
= −

2
2m
d
2
dx
2
The integral becomes:
< KE > =
_
1
0
Be
−x
_
−

2
2m
d
2
dx
2
_
Be
−x
dx
= −

2
2m
B
2
_
1
0
e
−2x
dx
= −

2
4m
B
2
_
e
−2
−1
¸
where B is the normalization constant found in problem 12.
Copyright © 2010 by Elsevier, Inc. All rights reserved.
28 3. OPERATORS AND WAVES - SOLUTIONS
Copyright © 2010 by Elsevier, Inc. All rights reserved.
4
The Hydrogen Atom -
Solutions
2. Per equation 4.9, the radial probability density for the 1s state hydrogen
is found by:
|P
1s
(r)|
2
=
_
2
_
1
a
0
r
a
0
e
r/a
0
_
2
=
2r
2
a
3
0
e
−2r/a
0
The 1s radial wave function used here is listed in table 4.2 in the row corre-
sponding to n=1 and l=0. Since hydrogen is the atom of interest, we used the
appropriate atomic number Z=1. We shall maximize the above expression
by taking the ﬁrst derivative and equating to zero:
d
dr
|P
1s
(r)|
2
=
4r
a
3
0
e
−2r/a
0
+
2r
2
a
3
0
_
−2
a
0
_
e
−2r/a
0
= 0
=
4r
a
3
0
e
−2r/a
0
−
4r
2
a
4
0
e
−2r/a
0
= 0
Divide both sides by e
−2r/a
0
and solve for r:
4r
a
3
0
−
4r
2
a
4
0
= 0
4r
a
3
0
_
1 −
r
a
0
_
= 0
Therefore we have one solution at r = 0 and one at r = a
0
. Let’s test the
intervals (0, a
0
) and (a
0
, ∞) to reveal the nature of these points. Let r = a
0
/2
29
Copyright © 2010 by Elsevier, Inc. All rights reserved.
30 4. THE HYDROGEN ATOM - SOLUTIONS
for the former and r = 2a
0
for the latter:
d
dr
|P
1s
(r)|
2
¸
¸
¸
¸
x=a
0
/2
=
4
a
3
0
_
a
0
2
_
e
−1
−
4
a
3
0
_
a
2
0
4
_
e
−1
= e
−1
_
2
a
2
0
−
1
a
2
0
_
> 0
d
dr
|P
1s
(r)|
2
¸
¸
¸
¸
x=2a
0
=
4
a
3
0
(2a
0
) e
−4
−
4
a
3
0
_
4a
2
0
_
e
−4
=
8
a
2
0
e
−4
(1 −2) < 0
Since the ﬁrst derivative of the radial probability density takes a positive val-
ues for (0, a
0
) and negative values for (a
0
, ∞), we may conclude that r = a
0
is in fact the maximum value. This is consistent with the description of the
Bohr radius as the innermost orbital radius of the electron in the hydrogen
atom.
4. Starting with equations 4.7 and 4.8,
dP = |ψ(r)|
2
r
2
sin(θ) dr dθ dψ
We construct our wavefunction by equation 4.4:
ψ(r, θ, φ) =
P
nl
(r)
r
Θ
lm
(θ) Φ
m
(θ, φ)
P
nl
(r), Θ
lm
(θ), and Φ
m
(θ, φ) for the 1s state may be found in tables 4.1 and
4.2 in rows corresponding to n=1, l=0, and m
l
=0 (Note that m
l
=0 is the only
possibility for l=0 thus only one row for l=0 appears in table 4.1). Carrying
on,
ψ(r, θ, φ) =
2
r
_
1
a
0
_
r
a
0
_
e
−r/a
0
1
√
2
1
√
2π
=
1
a
3/2
0
e
−r/a
0
1
√
π
Copyright © 2010 by Elsevier, Inc. All rights reserved.
31
We now integrate to ﬁnd the probability:
P =
_
|ψ(r, θ, φ)|
2
r
2
sin(θ) dr dθ dφ
=
1
π
1
a
3
0
_
a
0
0
r
2
e
−2r/a
0
dr
_
π
0
sin(θ) dθ
_
2π
0
dφ
=
_
a
0
0
4r
2
a
3
0
e
−2r/a
0
dr
Grouping 1/a
0
with r and dr:
P = 2
_
a
0
0
_
2r
a
0
_
2
e
−2r/a
0
d
_
r
a
0
_
Substitute ρ =
r
a
0
:
P = 2
_
1
0
ρ
2
e
−2ρ
dρ
This integral may be solved analytically using integration by parts:
_
1
0
ρ
2
e
−2ρ
dρ = −
1
2
e
−2ρ
ρ
2
¸
¸
¸
¸
1
0
−
_
1
0
2ρ
_
−
1
2
e
−2ρ
_
dρ
= −
1
2
e
−2
+
_
−
1
2
_
e
−2ρ
ρ
¸
¸
¸
¸
1
0
−
_
1
0
_
−
1
2
_
e
−2ρ
dρ
=
1
4
_
1 −5e
−2
_
5. For a given n-state, the possible values of l are n-1, n-2, . . . 0. Therefore,
n=5 allows l=0,1,2,3,4. For any given l value, the possiblities for the mag-
netic quantum number are m
l
= -l, -l+1, . . . 0, 1, . . . l. The results are as
follows:
l = 0 : m = 0
l = 1 : m = −1, 0, 1
l = 2 : m = −2, −1, 0, 1, 2
l = 3 : m = −3, −2, −1, 0, 1, 2, 3
l = 4 : m = −4, −3, −2, −1, 0, 1, 2, 3, 4
Copyright © 2010 by Elsevier, Inc. All rights reserved.
32 4. THE HYDROGEN ATOM - SOLUTIONS
7. Recall from equation 4.4 our construction of a wave function as a product
of radial and angular parts:
ψ(r, θ, φ) =
P
nl
(r)
r
Θ
lm
l
(θ)Φ(φ)
Our strategy will be to express the given wave function in terms of radial and
angular parts, then use their properties to check whether the wavefunction
is an eigenfunction of the Schrodinger equation and ﬁnd the corresponding
energy. For our given wavefunction, we see that we may separate it in terms
of dependence on θ and dependence on r as cos(θ) and r
2
e
−Zr/2a
0
. Note
that the presence of r
2
is due to fact the radial part is to be divided by r in
eq. 4.4 to construct the total wave function. We need not incorporate the
constant C as it will not aﬀect whether the wave function is a solution of the
Schrodinger equation. For the angular part, note that in table 4.1, cos(θ)
appears in the spherical harmonic corresponding to l = 1 and m
l
= 0.
When we distribute the left-hand side of eq. 4.4, we get:
−
2
2m
d
2
dr
2
P
nl
(r) +

2
l(l + 1)
2mr
2
P
nl
(r) −
1
4π
0
Ze
2
r
P
nl
(r)
Taking the second derivative of the radial wave function:
d
2
dr
2
P
nl
(r) = 2e
−Zr/2a
0
−
2rZ
a
0
e
−Zr/2a
0
+
Z
2
r
2
4a
0
e
−Zr/2a
0
Upon substitution and using l = 1, the left-hand side of eq. 4.4 becomes:
−
2
2m
d
2
dr
2
P
nl
(r) +

2
l(l + 1)
2mr
2
P
nl
(r) −
1
4π
0
Ze
2
r
P
nl
(r)
= −

2
m
e
−Zr/2a
0
+

2
rZ
ma
0
e
−Zr/2a
0
+

2
Z
2
r
2
8ma
0
e
−Zr/2a
0
+

2
m
e
−Zr/2a
0
−
Ze
2
r
4π
0
e
−Zr/2a
0
=

2
rZ
ma
0
e
−Zr/2a
0
−
Ze
2
r
4π
0
e
−Zr/2a
0
+

2
Z
2
r
2
8ma
0
e
−Zr/2a
0
Recall from equation 1.20 that the Bohr radius a
0
may expressed as follows:
a
0
=
4π
0

2
me
2
Copyright © 2010 by Elsevier, Inc. All rights reserved.
33
If we substitute this expression for a
0
into the ﬁrst term of the previous
equation, the ﬁrst two terms cancel and equation 4.5 becomes:

2
Z
2
8ma
0
r
2
e
−Zr/2a
0
= EP
nl
(r)
The radial wave function is in fact an eigenfunction of the Schrodinger equa-
tion with corresponding eigenvalue (energy):
E =

2
Z
2
8ma
0
9. We shall start with eq. 4.5:
_
−
2
2m
d
2
dr
2
+

2
l(l + 1)
2mr
2
−
1
4π
0
Ze
2
r
_
P
nl
(r) = EP
nl
(r)
And 1.20:
a
0
=
4π
0

2
me
2
The substitution as directed is ρ = r/a
0
, so we can directly substitute for F:
r = a
0
ρ. We let Z=1 as is the case hydrogen and carry out the substitution:
_
−
2
2m
d
2
d(ρa
0
)
2
+

2
l(l + 1)
2m(ρa
0
)
2
−
1
4π
0
Ze
2
ρa
0
_
P
nl
(r) = EP
nl
(r)
Since a
0
is a constant, we may bring it outside the derivative in the ﬁrst term
on the left-hand side and substitute eq. 1.20 for one a
0
in each of the ﬁrst
two terms.
_
−
1
a
0

2
2m
me
2
4π
0

2
d
2
dρ
2
+
1
a
0

2
l(l + 1)
2m(ρa
0
)
2
me
2
4π
0

2
−
1
a
0
Ze
2
ρa
0
_
P
nl
(r) = EP
nl
(r)
After making the appropriate cancellations and factoring we have:
e
2
a
0
1
4π
0
_
−
1
2
d
2
dρ
2
+
1
2
l(l + 1)
ρ
2
−
1
ρ
_
P
nl
(r) = EP
nl
(r)
_
−
1
2
d
2
dρ
2
+
l(l + 1)
2ρ
2
−
1
ρ
_
P
nl
(r) =
E
(1/4π
0
)(e
2
/a
0
)
P
nl
(r)
Copyright © 2010 by Elsevier, Inc. All rights reserved.
34 4. THE HYDROGEN ATOM - SOLUTIONS
12. The transition integral for the general case is given by
_
φ
∗
n
1
l
1
m
1
l
_
z
a
0
_
φ
n
2
l
2
m
2
l
dV
We are interested in the transition 3d
1
→ 2p
1
, so the transition integral
becomes:
_
φ
∗
3,2,1
_
z
a
0
_
φ
2,1,1
dV
From here on, we assume that the radial distance will be given in terms
relative to a
0
, so we take a
0
= 1 and constructs the wavefunctions by tables
4.1 and 4.2.
φ
3,2,1
=
P
3,2
(r)
r
Y
2,1
(θ, φ)
=
2
√
2
81
√
15
r
2
e
−r/3
_
−
_
15
8π
sin(θ)cos(θ)e
iφ
_
φ
2,1,1
=
P
2,1
(r)
r
Y
1,1
(θ, φ)
=
1
2
√
6
re
−r/2
_
−
_
3
8π
sin(θ)e
iφ
_
Now, we take the complex conjugate of φ
3,2,1
, substitute z = rcos(θ), and
proceed with the integration:
_
φ
∗
3,2,1
zφ
2,1,1
dV =
1
648π
_
r
4
e
−5r/6
sin
2
(θ)cos
2
(θ) dV
=
1
648π
_
∞
0
_
π
0
_
2π
0
r
6
e
−5r/6
sin
3
(θ)cos
2
(θ) dr dθ dφ
The integration may be separated as follows:
1
648π
_
∞
0
r
6
e
−5r/6
dr
_
π
0
sin
3
(θ)cos
2
(θ) dθ
_
2π
0
dφ =
1
648π
40310784
15625
4
15
2π
= 2.123
Copyright © 2010 by Elsevier, Inc. All rights reserved.
35
13. In order to ﬁnd the transition coeﬃcient for hydrogen, we will use the
results found in Appendix FF (online) along with eqs. 4.20 and 4.21. We
According to the selection rules, summarized in table 4.3, ∆m
l
= ±1 for x-
and y-polarized light. Therefore, we must calculate the transition integrals
for both the 2p
1
→1s
0
and 2p
−1
→1s
0
cases. As directed in Appendix FF,
we use the following in place of the z operator in eq. 4.21:
x =
1
2
(r
+
+r
−
)
y =
1
2i
(r
+
−r
−
)
We ﬁrst solve the case for x-polarized light.
I
12
= 2
_
|φ
∗
1s
0
xφ
2p
1
|
2
+|φ
∗
1s
0
xφ
2p
−1
|
2
_
Which gives us the following when we substitute the above expression for x:
φ
∗
1s
0
xφ
2p
1
=
1
2
(φ
∗
1s
0
r
+
φ
2p
1
+φ
∗
1s
0
r
−
φ
2p
1
)
φ
∗
1s
0
xφ
2p
−1
=
1
2
(φ
∗
1s
0
r
+
φ
2p
−1
+φ
∗
1s
0
r
−
φ
2p
−1
)
Since φ
∗
1s
0
r
+
φ
2p
1
and φ
∗
1s
0
r
−
φ
2p
−1
will equal zero, we may use the results given
in Appendix FF for the remaining terms, giving us:
I
12
= 2
_
1
6
R
2
i
+
1
6
R
2
i
_
=
2
3
R
2
i
Using the results of eqs. 4.23 and 4.26 into eq. 4.22:
A
21
=
6.078 ×10
15
(121.6)
3
1.109
6
= 6.25 ×10
8
per atom per second
We may follow the same procedure for y-polarized light. First, the transition
integral:
I
12
= 2
_
|φ
∗
1s
0
yφ
2p
1
|
2
+|φ
∗
1s
0
yφ
2p
−1
|
2
_
Copyright © 2010 by Elsevier, Inc. All rights reserved.
36 4. THE HYDROGEN ATOM - SOLUTIONS
Using the operator substitution:
φ
∗
1s
0
yφ
2p
1
=
1
2i
(φ
∗
1s
0
r
+
φ
2p
1
−φ
∗
1s
0
r
−
φ
2p
1
)
φ
∗
1s
0
yφ
2p
−1
=
1
2i
(φ
∗
1s
0
r
+
φ
2p
−1
−φ
∗
1s
0
r
−
φ
2p
−1
)
Once again, the nonzero terms lead to:
I
12
= 2
_
−1
6
R
2
i
+
−1
6
R
2
i
_
=
−2
3
R
2
i
The result for y-polarized light is then:
A
21
=
6.078 ×10
15
(121.6)
3
−1.109
6
= −6.25 ×10
8
per atom per second
15. a) Let’s imagine that the angular momentum vector l points from the
origin and lies on the surface of a cone, then the radius of the circle at the
base of the cone is:
r = |l|sinθ
And, for a change of the azimuthal angle dφ, the distance that the tip of the
angular momentum vector moves is given by:
d|l| = |l|sinθ dφ
b) We start with the relationship of the torque to angular to momentum:
d|l|
dt
= |τ|
This equation together with eq. 4.40 immediately leads to the equation:
d|l|
dt
=
e
2m
|l ×B|
The deﬁnition of the vector product then gives:
d|l|
dt
=
e
2m
|l|B sin(θ)
Copyright © 2010 by Elsevier, Inc. All rights reserved.
37
Substituting d|l| = |l|sinθ dφ, we may then ﬁnd our Larmor frequencies:
|l|sin(θ)
dφ
dt
=
e
2m
|l|B sin(θ)
ω
L
=
dφ
dt
=
eB
2m
17. The total angular momentum of the 4f electron will contain contribu-
tion form both the electron’s spin and its orbital angular momentum. Section
4.3.4 shows that the total angular momentum may take on the values:
J = j
1
+j
2
, j
1
+j
2
−1, . . . |j
2
−j
1
|
We let j
1
be our orbital angular momentum and j
2
the spin. The electron
occupies an f-state, therefore j
1
= 3, and the intrinsic spin of the electron
means j
2
= 1/2. We may now list the possible values for J:
J =
7
2
,
5
2
The spin-orbit coupling energy is then found by:
E
s−o
=
ζ
2
2
l =
3ζ
2
2
for j = l +
1
2
E
s−o
=
−ζ
2
2
(l + 1) = −2ζ
2
for j = l −
1
2
19. Using the two equations following eq. 4.50, the separation in energies
due to spin-orbit coupling is:
∆E =
ζ
2
2
l −
_
−ζ
2
2
_
(l + 1) =
_
ζ
2
2
_
(2l + 1)
Since we are using the atomic system of units, = 1 and
∆ =
ζ
2
(2l + 1) (1)
Copyright © 2010 by Elsevier, Inc. All rights reserved.
38 4. THE HYDROGEN ATOM - SOLUTIONS
We then let l=1 (due to the occupation of a p-orbital):
∆E =
3
2
ζ = 7.9 ×10
−6
ζ = 5.27 ×10
−6
The spin-orbit constant ζ for the levels neon may be calculated directly by
eq. 4.48, however we may use equation 4.51 to set up a relationship between
ζ
He
and ζ
Ne
. Since He
+
and Ne
9+
are both hydrogenic atoms (containing
only one electron), the quantity
_
_
a
0
Zr
_
3
_
should be equal in both cases as the grouping of a
0
and z with r will account
for the scaling of the radial distance with changing nuclear charge. We now
use the appropriate atomic numbers of helium and neon to generate a ratio
between ζ
He
and ζ
Ne
.:
ζ
Ne
ζ
He
=
10
4
2
4
= 625 ⇒ζ
Ne
= 625ζ
He
= 3.29 ×10
−3
For the separation between the 3p
3/2
and 3p
1/2
levels, we use equation (1)
above:
∆E =
3.29 ×10
−3
2
(2l + 1) = 4.94 ×10
−3
A comparison with the separation of the corresponding helium states reveals
the vastly increased impact of spin-orbit coupling in the case of neon.
Copyright © 2010 by Elsevier, Inc. All rights reserved.
5
Many-Electron Atoms -
Solutions
1. An electron has an intrinsic spin s = 1/2, therefore the possible val-
ues of m
s
are ±1/2. The 4f state of the electron has an orbital angular
momentum quantum number l = 3 and therefore may have values m
l
=
−3, −2, −1, 0, 1, 2, 3.
In the case of two electrons occupying the 4f state, we ﬁrst see from our
results above, that there are 7 possible m
l
quantum numbers 2 m
s
quantum
numbers. There are thus 14 possible states in which to put the ﬁrst electron.
Once we assign ﬁrst electron, there are now 13 possible states. Because the
order in which placed them does not matter in our conﬁguration, we then
divide the product by two:
# of distinct states =
14 · 13
2
= 91
3. Using the form of eq. 5.6, we attribute each row of the determinant
to a particular state and each column to a particular electron occupying a
corresponding state (row). Our system contains these electrons, therefore
N=3 in our coeﬃcient and
Φ =
_
1
3
¸
¸
¸
¸
¸
¸
ψ
10
α(1) ψ
10
α(2) ψ
10
α(3)
ψ
10
β(1) ψ
10
β(2) ψ
10
β(3)
ψ
20
α(1) ψ
20
α(2) ψ
20
α(3)
¸
¸
¸
¸
¸
¸
39
Copyright © 2010 by Elsevier, Inc. All rights reserved.
40 5. MANY-ELECTRON ATOMS - SOLUTIONS
4. For each of these elements, we shall ﬁll their shells in the following order:
1s, 2s, 2p, 3s, 3p, 4s, 4p, and so on. The number of electrons in a neutral
atom of any element will equal the number of protons in its nucleus, therefore
we use the atomic number of each element as the number of electrons we use
to ﬁll these shells:
Flourine(Z = 9) : 1s
2
2s
2
2p
5
Magnesium(Z = 12) : 1s
2
2s
2
2p
6
3s
2
[Ne] 3s
2
Silicon(Z = 14) : 1s
2
2s
2
2p
6
3s
2
3p
2
[Ne] 3s
2
3p
2
Potassium(Z = 19) : 1s
2
2s
2
2p
6
3s
2
3p
6
4s
1
[Ar] 4s
1
Cobalt(Z = 27) : 1s
2
2s
2
2p
6
3s
2
3p
6
4s
2
3d
7
[Ar] 4s
2
3d
7
An alternate (shorter) way to notate these, as I’ve done above to the right,
is to use the symbol of the inert gas that occurs closest before a particular
element to denote the ﬁlled shells through complete rows of the periodic table
and merely denote the ﬁlling of the remaining shells.
7. The nitrogen atom will have the following ground conﬁguration: 1s
2
2s
2
2p
3
.
Since the 1s and 2s orbitals have less energy than the 2p orbital, the next
higher conﬁgurations are 1s
2
2s
2
2p
2
3s and 1s
2
2s
2
2p
2
3p.
8. ns
2
: Here are two electrons with l=0, therefore, L=0. Their spins pro-
duce S=0,1. Since L+S must be even for two equivalent electrons, the only
possibility is
1
S (S=0, L=0).
nd
2
: Two electrons with l=2 may produce L=4,3,2,1,0. Again, the addi-
tion of their spin angular momenta produce S=0,1. Imposing the condition
that L+S must be even, we have the following conﬁgurations:
1
G (L=4,
S=0),
3
F(L=3, S=1),
1
D (L=2, S=0),
3
P(L=1, S=1), and
1
S (S=0, L=0).
4f
2
: These two electrons occupy f-states (l=3), thus may contribute to
L=6,5,4,3,2,1,0. As has been the case above, S=0,1. Ipmosing the condi-
tion that L+S must be even, we have the following LS terms:
1
I (L=6,
S=0),
3
H (L=5, S=1),
1
G (L=4, S=0),
3
F (L=3, S=1),
1
D (L=2, S=0),
3
P(L=1, S=1), and
1
S (S=0, L=0).
Let’s apply Hund’s rules to the 4f conﬁguration. Of these terms, the ones
Copyright © 2010 by Elsevier, Inc. All rights reserved.
41
corresponding to the maximum value of S are
3
H,
3
F, and
3
P. Of these, the
one with maximum value of L (
3
H) should lie lowest.
11. It is important to take note here that we have a case of non-equvalent
electrons for the [Xe] 4f 5d conﬁguration and one of equivalent electrons
with [Xe] 4f
2
. For [Xe] 4f 5d, one electron has l=3 and the other l=2.
Since there is no additional constraint on the values of L and S except for
the rules of addition of angular momentum, L may take L=5,4,3,2, or1 and
S=0,1.
For 4f
2
, we must impose the condition that L+S be even. The two l=3
electrons could form L=6,5,4,3,2,1,0 and S=0,1. The allowed LS terms are
then (L=6, S=0), (L=5, S=1), (L=4, S=0), (L=3, S=1), (L=2, S=2), (L=1,
S=1), and (L=0,S=0).
13. As described in section 5.5.2, the nuclear charge increases as we go
across a row in the periodic table, thus decreasing the average value of the
electrons’ distance from the nucleus. This causes an increase in the Coulomb
interaction, thus we expect sulfur (S) to possess the greatest Coulomb inter-
action of these three elements.
15. Recall from equation 4.51 that the strength of the spin-orbit interac-
tion increases with atomic number as Z
4
. Since the spin-orbit interaction is
mainly responsible for the breakdown of LS coupling, the O atom has the
smallest spin-orbit interaction and the states of O corresponds most nearly
correspond to pure LS terms.
17. To ﬁnd the total energy, we simply click on neon in the periodic ta-
ble interface, then the red arrow on the bottom row of controls to run the
calculation. Then, click on the ”Averages” tab to display the kinetic, po-
tential, and total energies. We see a value of -128.55 for the total energy.
We now remove a 2p electron by selecting the 2p shell via the incremental
objects in the bottom row of controls, then clicking the (-e) button. Verify
you have removed the electron from the correct shell as the conﬁguration in
the second to last row of controls should now read 1s
2
2s
2
2p
5
. Now, run the
applet again for the neon ion and ﬁnding the energy under the ”Averages”
tab, you should ﬁnd the energy to be approximately 127.82. The binding
Copyright © 2010 by Elsevier, Inc. All rights reserved.
42 5. MANY-ELECTRON ATOMS - SOLUTIONS
energy is the diﬀerence of these two values:
|−128.55 −(−127.82)| ≈ .73
Going back to the neutral neon atom, the third column to the left on the page
obtained after striking the Averages tab gives a 2p single electron energy of
0.85. This is very close to the diﬀerence in total energies of the neutral Ne
atom and the ion.
19. We can calculate the Slater integrals for Si, P, and S by ﬁrst select-
ing the appropriate element in the periodic table of the Hartree-Fock applet,
clicking the red arrow to run the calculation, then clicking the ”Other” tab.
Now, the second box should be changed to ”2”, then continuing across: ”3”,
”p”, ”3”, ”p”. The red arrow on the right runs the calculation. We obtain
the following values.
F
2
(3p, 3p)
Element au cm
−1
Si .166 36418
P .197 43180
S .220 48323
We see a deﬁnite increase along the row, constitent with the qualitative
result obtained in problem 13.
21. Using ﬁg. 5.5, the energy of
1
S of the 1s 3s conﬁguration is approxi-
mately 185,000 cm
−1
while
1
P has an energy of approx. 170,000 cm
−1
. The
energy of the emitted photon should equal the diﬀerence of the energy levels,
thus:
∆E = 25, 000 cm
−1
According to equation 5.12, 1 eV of photon energy corresponds to light of
wave number 8065.54 cm
−1
. The photon energy emitted in this transition
must equal:
25, 000 cm
−1
8065.54 cm
−1
/eV
= 3.10 eV
Using the relation E = hc/λ,
λ =
hc
E
=
1240 eV · nm
3.10 eV
= 400 nm
Copyright © 2010 by Elsevier, Inc. All rights reserved.
6
The Emergence of Masers and
Lasers - Solutions
1. As given in ﬁgure 6.7, the range of the
4
F
2
band is approximately
17, 000 cm
−1
to 19, 000 cm
−1
. As these quantities are given in terms of
1/λ, we may quickly obtain the corresponding frequencies in Hz by using the
relation I.22:
c = λf ⇒f =
c
λ
Therefore, we may convert c to cm/s and simply multiply:
17 ×10
3
cm
−1
_
3 ×10
10
cm
s
_
= 5.1 ×10
14
Hz
19 ×10
3
cm
−1
_
3 ×10
10
cm
s
_
= 5.7 ×10
14
Hz
Similarly, for
4
F
1
:
23 ×10
3
cm
−1
_
3 ×10
10
cm
s
_
= 6.9 ×10
14
Hz
27 ×10
3
cm
−1
_
3 ×10
10
cm
s
_
= 8.1 ×10
14
Hz
3. The 2p core is given to be in the state
2
P
3/2
. Using the following scheme
for spectroscopic notation:
2s+1
L
J
43
Copyright © 2010 by Elsevier, Inc. All rights reserved.
44 6. THE EMERGENCE OF MASERS AND LASERS - SOLUTIONS
we take J = 3/2. Since the excited electron (3p) has orbital angular momen-
tum l=1, the possible values for K are:
K = J +l, J +l −1, . . . |J −l| =
5
2
,
3
2
,
1
2
The spin orbit interaction spits each level with a value of K into two levels
corresponding to the total angular momentum having the values K ±
1
2
.
5. The situation described in problem 4 gives I = 1/2 and the fact that
we are dealing with hydrogen which means we have one electron, for which
S = 1/2. The possible values for F are then F = 0, for which M
F
= 0;
and F = 1, for which M
F
= −1, 0, 1. Since the F=0 level is not split by a
magnetic ﬁeld, we only calculate g
F
for F = 1.Using eq. 6.17,
For F=1, g
F
=
(1 + 1) −
1
2
(
1
2
+
1
2
) +
1
2
(
1
2
+ 1)
1 + 1
=
9
8
Therefore, we consider the F=1 states via equation 6.5. Since g
F
is positive,
the M
F
= 1 state will have a negative µ
z
value and be drawn to regions of
low magnetic ﬁeld. Consequently, the M
F
= −1 will have a positive µ
z
value
and be drawn to regions of high magnetic ﬁeld outside the magnetic trap.
These atoms will be lost, decreasing the kinetic energy and temperature of
the atoms caught in the trap.
7. Given S = 1/2 and I = 5/2, our possible values for F are F = 3 and
F = 2. These levels split as follows:
F = 3 : M
F
= −3, −2, −1, 0, 1, 2, 3
F = 2 : M
F
= −2, −1, 0, 1, 2
As we see, the M
F
= −3 and M
F
= 3 states are not mixed with any states
where F = 2, thus will be “pure” states with straight lines. All other states
of F = 3 and all states where F = 2 will be mixed, thus will have curved
lines.
Copyright © 2010 by Elsevier, Inc. All rights reserved.
7
Statistical Physics - Solutions
1. As described in the text, there are sixteen possible outcomes of ﬂipping
a coin four times. The possible outcomes can be classiﬁed according to the
number of heads. We may use equation 7.2 (or 7.3) to calculate the statisti-
cal weight associated with each outcome by letting N = 4 and n equal the
number of heads obtained in each distribution:
HHHH :
_
4
4
_
=
4!
4!0!
= 1
HHHT :
_
4
3
_
=
4!
3!1!
= 4
HHTT :
_
4
2
_
=
4!
2!2!
= 6
HTTT :
_
4
1
_
=
4!
1!3!
= 4
TTTT :
_
4
3
_
=
4!
0!4!
= 1
3. We may set up equation 7.7 for each energy level:
n
2
g
2
=
n
2
2
=
N
Z
e

2
/k
B
T
(1)
45
Copyright © 2010 by Elsevier, Inc. All rights reserved.
46 7. STATISTICAL PHYSICS - SOLUTIONS
n
1
g
1
=
n
1
1
=
N
Z
e

1
/k
B
T
(2)
Divide (1) by (2):
n
2
n
1
= 2e
−(
2
−
1
)/k
B
T
We are given
2
−
1
= 0.025 eV , T = 298K, and we use k
B
= 8.617 ×
10
−5
eV/K:
n
2
n
1
= 2 exp
_
−0.025 eV
(8.617 ×10
−5
eV/K)(298 K)
_
= .755
5. F
1
(u) is deﬁned in conjunction with Fig. 7.2:
F
1
(u) =
4
√
π
u
2
e
−u
2
We maximize F
1
(u) by setting the derivative of F
1
(u) with respect to u to
zero:
dF
1
du
=
4
√
π
_
2ue
−u
2
−2u
3
e
−u
2
_
= 0
2ue
−u
2
(1 −u
2
) = 0
u = 0, ±1
We test the intervals (0, 1) and (1, ∞):
F

1
_
1
2
_
> 0 and F

1
(2) < 0
Thus, F
1
(u) has a maximum at u = 1.
7. To ﬁnd the average value of v
2
, we calculate the integral:
(v
2
)
av
=
_
∞
0
v
2
P(v) dv
Use equation 7.22 for P(v)dv:
(v
2
)
av
=
_
∞
0
4πv
4
_
m
2πk
B
T
_
3/2
e
−mv
2
/2k
B
T
dv
Copyright © 2010 by Elsevier, Inc. All rights reserved.
47
We may use eq. F.2 in Appendix F to ﬁnd I
4
(1) = 3
√
π/8, thus (v
2
)
av
=
3k
B
T/m.
9. To ﬁnd
rms
, we need to ﬁrst ﬁnd (
2
)
av
, then take the square root.
Analogous to example 7.3,
(
2
)
av
=
_
∞
0

2
P() d
Using equation 7.26:
(
2
)
av
=
2
√
π
_
∞
0

5/2
(k
B
T)
3/2
e
/k
B
T
d
Per equation 7.27, u = /k
B
T and we may write the above equation in terms
of u:
(
2
)
av
=
2
√
π
(k
B
T)
2
_
∞
0
u
5/2
e
−u
du
At this stage, we can work the integral according to the procedure expressed
in eqs. F.8-F.14:
(
2
)
av
=
2
√
π
(k
B
T)
2
Γ(7/2)
=
2
√
π
(k
B
T)
2
Γ(7/2)
=
2
√
π
(k
B
T)
2
5
2
Γ(5/2)
=
2
√
π
(k
B
T)
2
5
2
3
2
Γ(3/2)
=
2
√
π
(k
B
T)
2
5
2
3
2
1
2
Γ(1/2)
=
15
4
(k
B
T)
2
We now take the square root of this last quantity:

rms
=
√
15
2
k
B
T
Copyright © 2010 by Elsevier, Inc. All rights reserved.
48 7. STATISTICAL PHYSICS - SOLUTIONS
11. According to the equipartition of energy (end of section 7.3.1), each
particle has an average kinetic energy of k
B
T/2 for each degree of freedom.
For three degrees of freedom:

av
=
3
2
k
B
T
Therefore, total kinetic energy of a mole of gas should be:
=
3
2
k
B
TN
A
=
3
2
(1.381×10
−23
J/K)(298 K)(6.022×10
23
particles) = 3717 J
13. Per equation 7.28,
u(f) =
8πf
2
c
3
_
hf
e
hf/k
B
T
−1
_
We apply the product rule and quotient rule of diﬀerentiation:
du(f)
df
=
16πf
c
3
_
hf
e
hf/k
B
T
−1
_
+
8πf
2
c
3
_
h
_
e
hf/k
B
T
−1
_
−
h
2
f
k
B
T
e
hf/k
B
T
(e
hf/k
B
T
−1)
2
_
Let’s set this expression equal to zero and multiply both sides by
_
e
hf/k
B
T
−1
_
2
:
16πf
2
c
3
_
e
hf/k
B
T
−1
_
+
8πf
2
c
3
_
e
hf/k
B
T
−1
_
−
8πf
2
c
3
_
hf
k
B
T
e
hf/k
B
T
_
= 0
Multiply both sides by c
3
/8πhr
2
and let x = hf/k
B
T, then simplify:
2 (e
x
−1) + (e
x
−1) −xe
x
= 0
3e
x
−xe
x
−3 = 0 ⇒(3 −x)e
x
= 3
15. Equation 7.28 gives the energy density as a function of frequency:
u(f) =
8πf
2
c
3
_
hf
e
hf/k
B
T
−1
_
Copyright © 2010 by Elsevier, Inc. All rights reserved.
49
We may derive an expression as a function of wavelength by using relation
I.23 from the introductory chaper: f = c/λ.
u(λ) =
8π(c/λ)
2
c
3
_
h(c/λ)
e
hc/λk
B
T
−1
_
=
8π
λ
3
_
h
e
hc/λk
B
T
−1
_
19. According to eq. 7.50, the work done on the gas is given by:
dW = −P dV
Per the ideal gas law, we solve for P:
P =
nRT
V
We then integrate to ﬁnd W and change sign to reﬂect work done by the gas:
W = nRT
_
V
3
V
2
dV
V
= nRT ln
_
V
3
V
2
_
Eq. 7.48 states dE = dQ
to
+ dW
on
. As found above, the work done by the
gas is equal to nRT ln(V
3
/V
2
). If the energy remains constant, the amount
of heat added to the gas must be equal to the work done by the gas and
dQ
to
= nRT ln
_
V
3
V
2
_
21. We ﬁrst solve eq. 7.66 for T
c
:
1
2.612
_
N
V
_
=
_
2πmk
B
T
c
h
2
_
3/2
T
c
=
_
1
2.612
N
V
_
2/3
h
2
2πmk
B
Substitute our given density:
N
V
= 5.0 ×10
14
atoms/cm
3
Copyright © 2010 by Elsevier, Inc. All rights reserved.
50 7. STATISTICAL PHYSICS - SOLUTIONS
T
c
=
_
1.16 ×10
9
_
h
2
πmk
B
=
7.31 ×10
−17
m
We may supply the appropriate atomic mass to calculate the critical tem-
perature. This equation may appear a bit peculiar due to a quantity of
temperature appearing on the left-hand side and one of mass on right. We
can take a closer look at the units to more clearly see this relationship. For
clarity, let’s use SI units of J, K, m, and kg for the necessary quantities:
T
c
=
_
1
2.612
N
V
_
2/3
h
2
2πmk
B
⇒[T
c
] =
1
m
2
J
2
· s
2
kg · J/K
=
1
m
kg m/s
2
· s
2
· K
kg
= K
23. Equation 7.76 gives the probability that an energy will fall between
and +d:
P() d =
3
2

−3/2
F

1/2
d
1
e
(−
F
)/k
B
T
+ 1
Thus, we may substitute our values for
F
and T then carry out the integration
numerically:
At T = 295K :
3
2
_
4.1
3.9
3
−3/2

1/2
1
e
(−
F
)/295k
B
)
+ 1
d = 5.9 ×10
−17
At T = 3000K :
3
2
_
4.1
3.9
3
−3/2

1/2
1
e
(−
F
)/3000k
B
)
+ 1
d = 5.9×10
−17
= 2.41×10
−3
It is seen here that the electron is much more likely to be found with an
energy between 3.9 ecV and 4.1 eV at 3000K.
25. We substitute our given quantities into the given formula, along with
n
c
= 1 since gold is a monovalent metal:
N
V
=
19.32 g/cm
3
(6.022 ×10
23
atoms/mole)
197 g/mole
Copyright © 2010 by Elsevier, Inc. All rights reserved.
51
N
V
= 5.90 ×10
22
electrons/cm
3
27. (a) Using the formula from problem 24 and the fact that n
c
= 1 since
sodium is a monovalent metal:
N
V
=
n
c
ρN
A
M
=
(.971 g/cm
3
)(6.022 ×10
23
atoms/mole)
23.0 g/mole
= 2.54 ×10
22
electrons/cm
3
(b) To ﬁnd the Fermi energy, we use equation 7.73 as well as the quantity
we just calculated in part (a):

F
=
h
2
2m
_
3
8π
N
V
_
2/3
=
(4.13 ×10
−15
)
2
2(5.48 ×10
−4
)
_
3
8π
(2.54 ×10
22
)
_
2/3
= 3.26 ×10
−12
eV
(c) The equation after example 7.6 relates the Fermi temperature to the
Fermi energy:

F
= k
B
T
F
⇒T
F
=

F
k
B
Now, we substitute the Boltzmann constant and the Fermi energy found in
part (c):
T
F
=
3 eV
8.617 ×10− =
−5
= 35, 418K
Copyright © 2010 by Elsevier, Inc. All rights reserved.
52 7. STATISTICAL PHYSICS - SOLUTIONS
Copyright © 2010 by Elsevier, Inc. All rights reserved.
8
Electronic Structure of Solids -
Solutions
1. The cesium choride lattice is depicted in Figure 8.5. If we take a lighter-
shaded sphere to represent a cesium ion, we see that there are 8 chloride ions
that serve as the nearest neighbors. Or, if the cesium ion serves as the center
point in a body-centered cubic lattice, the eight nearest neighbors are the
choloride ions at the corners.
Returning to Figure 8.5, recall that the designation of which shade cor-
responds to which ion is arbitrary. Therefore, now alllow the darker spheres
to be the cesium ions. By extrapolating the ﬁgure in one more direction, it
can be seen that cesium would have 6 next-nearest neighbors.
3. We refer to Fig. 8.8(b) where we will use vectors to denote the loca-
tions of the carbon atoms from a ﬁxed origin, then use vector relationships
to determine the bond angle. First, we select our origin to be located at point
A in the lattice. The other two point of interest are point B and the atom
located in the center of the top face shown. Our primitive
ˆ
i and
ˆ
j will be
oriented along the edges of the top face (starting at point A) and the
ˆ
k vector
will point directly upward from point A. We may now construct the position
vectors to each point. Let a be the position vector from the origin to the top
face-centered atom, while b points from point B to the origin (point A). It is
our intent to construct a vector c pointing from the top face-centered atom
to point B. We will then ﬁnd the angle between b and c Thus:
a =
1
2
ˆ
i +
1
2
ˆ
j, b = −
1
4
ˆ
i −
1
4
ˆ
j +
1
4
ˆ
k
53
Copyright © 2010 by Elsevier, Inc. All rights reserved.
54 8. ELECTRONIC STRUCTURE OF SOLIDS - SOLUTIONS
Our strategy wil be to ﬁnd the angle between b and c, so let’s ﬁrst ﬁnd c.
By inspection we see that:
c = a +b
We substitute the expressions for a and b:
c = a +b =
_
1
2
−
1
4
_
ˆ
i +
_
1
2
−
1
4
_
ˆ
j −
1
4
ˆ
k
=
1
4
ˆ
i +
1
4
ˆ
j −
1
4
ˆ
k
Now we use the dot (or scalar) product of b and c to ﬁnd the angle between
the two:
b · c = |b||c| cos(θ)
For the left-hand side:
b · c = −
1
16
−
1
16
+
1
16
= −
1
16
For the right-hand side:
|b||c| cos(θ) =
_
3
16
_
3
16
cos(θ) =
3
16
cos(θ)
Equating the two:
−
1
16
=
3
16
cos(θ)
cos(θ) = −
1
3
θ = cos
−1
_
−1
3
_
= 109
◦
28

5. For the simple cubic lattice, referring to ﬁgure 8.2, the nearest neigh-
bor of any point will be located along the length of a side of the cube. Thus,
the nearest neighbor distance is simply “a”. In the case of the face-centered
cubic lattice, Figure 8.6, for example, we may ﬁnd the nearest neighbor dis-
tance by ﬁnding the midpoint distance along the diagonal of a face. We may
simply ﬁnd the diagonal distance via the Pythagorean theorem:
√
a
2
+a
2
=
√
2a
Copyright © 2010 by Elsevier, Inc. All rights reserved.
55
The midpoint, thus nearest neighbor distance, is then (
√
2/2)a.
7. Our strategy to ﬁnd the center-to-center distance of the copper ions will
be to ﬁnd the volume of one cell, from which we will ﬁnd the length of an
edge of the cube, then use this to arrive at the face-centered nearest-neighbor
distance. First, we calculate the number of atoms per unit volume:
8.96 g/cm
3
63.5 g/mole
= .141
mole
cm
3
.141
mole
cm
3
_
6.022 ×10
23
atoms
mole
_
= 8.494 ×10
22
atoms
cm
3
Now, we wish to know the volume per unit cell. A face-centered cubic lattice
has four atoms per cell, therefore:
4 atoms/cell
8.494 ×10
22
atoms/cm
3
= 4.709 ×10
−23
cm
3
cell
One edge of the cube should now have length:
_
1.648 ×10
−22
_
1/3
= 3.611 ×10
−8
cm
As given in problem 5, the nearest-neighbor distance is then (
√
2/2)a, where
a is the length of an edge of the cube.
√
2
2
a = 2.553 ×10
−8
cm
9. We shall begin by constructing the position vectors of each of the corners,
starting with the corner directly above the origin indicated in Fig. 8.6 (b)
and continuing clockwise:
l
1
= a
ˆ
k
l
2
= a
ˆ
j +a
ˆ
k
l
3
= a
ˆ
i +a
ˆ
j +a
ˆ
k
l
4
= a
ˆ
i +a
ˆ
k
Copyright © 2010 by Elsevier, Inc. All rights reserved.
56 8. ELECTRONIC STRUCTURE OF SOLIDS - SOLUTIONS
The primitive vectors from eq. 8.4 are:
a
1
=
a
2
_
ˆ
j +
ˆ
k
_
, a
2
=
a
2
_
ˆ
k +
ˆ
i
_
, a
3
=
a
2
_
ˆ
i +
ˆ
j
_
Our strategy will be to solve these primitive vectors for
ˆ
i,
ˆ
j, and
ˆ
k and
substitute into the lattice vectors. We ﬁrst add a
1
and a
2
:
a
1
+a
2
=
a
2
(
ˆ
i +
ˆ
j) +a
ˆ
k
Then subtract a
3
:
a
1
+a
2
−a
3
= a
ˆ
k
We now solve for
ˆ
k:
ˆ
k =
a
1
+a
2
−a
3
a
We continue in such manner to ﬁnd:
ˆ
i =
a
2
+a
3
−a
1
a
ˆ
j =
a
1
−a
2
+a
3
a
We ﬁnally carry out the substitution:
l
1
= a
1
+a
2
−a
3
l
2
= a
1
−a
2
+a
3
+a
1
+a
2
−a
3
= 2a
1
l
3
= a
2
+a
3
−a
1
+a
1
−a
2
+a
3
+a
1
+a
2
−a
3
= a
1
+a
2
+a
3
l
4
= a
2
+a
3
−a
1
+a
1
+a
2
−a
3
= 2a
2
11. The results from problems 4 & 5 are as follows:
simple cubic : d = a ⇒R =
a
2
bcc : d =
√
3
2
a ⇒R =
√
3
4
a
fcc : d =
2
2
a ⇒R =
√
2
4
a
Copyright © 2010 by Elsevier, Inc. All rights reserved.
57
We now substitute these values of R into the packing fraction equation:
simple cubic : F = 1
4
3
π
_
a
3
_
3
a
3
=
π
6
bcc : F = 2
4
3
π
_
√
3a
4
_
3
a
3
=
√
3π
8
fcc : F = 4
4
3
π
_
√
2a
4
_
3
a
3
=
√
2π
6
13. Equation 8.2 gives the following bcc primitive vectors:
a
1
=
a
2
_
ˆ
j +
ˆ
k −
ˆ
i
_
, a
2
=
a
2
_
ˆ
k +
ˆ
i −
ˆ
j
_
, a
3
=
a
2
_
ˆ
i +
ˆ
j −
ˆ
k
_
We must use eqs. 8.18-8.20 to determine the primitive vectors of the corre-
sponding reciprocal lattice. However, it would be of great beneﬁt to calculate
the necessary cross-products ﬁrst. We shall use the determinant method here:
a
2
×a
3
=
a
2
4
¸
¸
¸
¸
¸
¸
ˆ
i
ˆ
j
ˆ
k
1 −1 1
1 1 −1
¸
¸
¸
¸
¸
¸
=
a
2
2
ˆ
j +
a
2
2
ˆ
k
a
3
×a
1
=
a
2
4
¸
¸
¸
¸
¸
¸
ˆ
i
ˆ
j
ˆ
k
1 1 −1
−1 1 1
¸
¸
¸
¸
¸
¸
=
a
2
2
ˆ
i +
a
2
2
ˆ
k
a
1
×a
2
=
a
2
4
¸
¸
¸
¸
¸
¸
ˆ
i
ˆ
j
ˆ
k
−1 1 1
1 −1 1
¸
¸
¸
¸
¸
¸
=
a
2
2
ˆ
i +
a
2
2
ˆ
j
Now we are ready to use eqs. 8.18-8.20 to calculate the primitive vectors of
the reciprocal lattice:
b
1
= 2π
a
2
3
_
ˆ
j +
ˆ
k
_
a
2
4
_
ˆ
j +
ˆ
k −
ˆ
i
_
·
_
ˆ
j +
ˆ
k
_ =
4π
a
_
ˆ
j +
ˆ
k
_
Copyright © 2010 by Elsevier, Inc. All rights reserved.
58 8. ELECTRONIC STRUCTURE OF SOLIDS - SOLUTIONS
b
2
= 2π
a
2
3
_
ˆ
i +
ˆ
k
_
a
2
4
_
ˆ
j +
ˆ
k −
ˆ
i
_
·
_
ˆ
j +
ˆ
k
_ =
4π
a
_
ˆ
i +
ˆ
k
_
b
3
= 2π
a
2
3
_
ˆ
i +
ˆ
j
_
a
2
4
_
ˆ
j +
ˆ
k −
ˆ
i
_
·
_
ˆ
j +
ˆ
k
_ =
4π
a
_
ˆ
i +
ˆ
j
_
Compared to eq. 8.4, we see that this reciprocal lattice is nothing but an
fcc-lattice with sides of length 8π/a.
15. We shall once again use eqs. 8.18-8.20 to calculate the primitive vectors
of the reciprocal lattice. But, again we carry out the cross-products ﬁrst:
a
2
×a
3
=
a
2
4
¸
¸
¸
¸
¸
¸
ˆ
i
ˆ
j
ˆ
k
1 0 1
1 1 0
¸
¸
¸
¸
¸
¸
=
a
2
4
_
−
ˆ
i +
ˆ
j +
ˆ
k
_
a
3
×a
1
=
a
2
4
¸
¸
¸
¸
¸
¸
ˆ
i
ˆ
j
ˆ
k
1 1 0
0 1 1
¸
¸
¸
¸
¸
¸
=
a
2
4
_
ˆ
i −
ˆ
j +
ˆ
k
_
a
1
×a
2
=
a
2
4
¸
¸
¸
¸
¸
¸
ˆ
i
ˆ
j
ˆ
k
0 1 1
1 0 1
¸
¸
¸
¸
¸
¸
=
a
2
4
_
ˆ
i +
ˆ
j −
ˆ
k
_
Since the denominator of each of eqs. 8.18-8.20 is the same, we may go ahead
and calculate it to substitute into each equation:
a
1
· (a
2
×a
3
) =
a
2
_
ˆ
j +
ˆ
k
_
·
a
2
4
_
−
ˆ
i +
ˆ
j +
ˆ
k
_
=
a
3
4
Now our reciprocal lattice primitive vectors are:
b
1
=
2π
a
_
−
ˆ
i +
ˆ
j +
ˆ
k
_
=
4π
a
1
2
_
−
ˆ
i +
ˆ
j +
ˆ
k
_
b
2
=
2π
a
_
ˆ
i −
ˆ
j +
ˆ
k
_
=
4π
a
1
2
_
ˆ
i −
ˆ
j +
ˆ
k
_
b
3
=
2π
a
_
ˆ
i +
ˆ
j −
ˆ
k
_
=
4π
a
1
2
_
ˆ
i +
ˆ
j −
ˆ
k
_
Copyright © 2010 by Elsevier, Inc. All rights reserved.
59
We factored out 1/2 so as to get us closer to the form of eq. 8.2. Notice that
these reciprocal vectors are nothing but the primitive vectors of a bcc-lattice
with sides of length 4π/a.
18. Once again, we shall use eqs. 8.18-8.20 extensively as we want to show
that:
a
1
= 2π
b
2
×b
3
b
1
· (b
2
×b
3
)
(1)
a
2
= 2π
b
3
×b
1
b
1
· (b
2
×b
3
)
(2)
a
1
= 2π
b
1
×b
2
b
1
· (b
2
×b
3
)
(3)
Where b
1
, b
2
, and b
3
are given by eqs. 8.18-8.20. Let us calculate the
cross-product b
2
×b
3
ﬁrst since it will be used in each equation.
b
2
×b
3
= 4π
2
_
a
3
×a
1
a
1
· (a
2
×a
3
)
_
×
_
a
1
×a
2
a
1
· (a
2
×a
3
)
_
We ﬁrst make the important observation that the denominators of the terms
on each side of the cross-product results in a scalar quantity. Therefore, it
can be brought outside the vector product and we may carry out the cross-
product. Let us proceed by writing just the operation in question:
(a
3
×a
1
) ×(a
1
×a
2
)
The following vector relation, which can be found in most books that discuss
vector calculus, can be of immense help here:
A×(B×C) = B(A· C) −C(A· B)
We relate the vectors that occur in the relation above to the vectors involved
in our desired operation as follows:
A = a
3
×a
1
B = a
1
C = a
2
After the substitution:
Copyright © 2010 by Elsevier, Inc. All rights reserved.
60 8. ELECTRONIC STRUCTURE OF SOLIDS - SOLUTIONS
(a
3
×a
1
) ×(a
1
×a
2
) = a
1
[(a
3
×a
1
) · a
2
] −a
2
[(a
3
×a
1
) · a
1
]
= a
1
[a
1
· (a
2
×a
3
)] +a
2
[(a
1
×a
1
) · a
3
]
To arrive at this ﬁrst term in the second expression of the right-hand side,
we used the following vector relation:
A· (B×C) = (A×B) · C
Notice that the part of this term in brackets is now identical to the expression
for the denominator. We also see that the second term is equal to zero since
we have a vector in a cross-product with itself. Our result for b
2
×b
3
is now:
b
2
×b
3
= 4π
2
a
1
We can use the same method for the other relevant cross-products to obtain
them as well:
b
3
×b
1
= 4π
2
a
2
b
1
×b
2
= 4π
2
a
3
The denominators of (1), (2), and (3), which are identical may be also readily
solved using the result from the cross-products above:
b
1
· (b
2
×b
3
) = 2π
a
2
×a
3
a
1
· (a
2
×a
3
)
· 4π
2
a
1
= 8π
3
By substituting the results we have obtained into the right-hand side of (1),
we see that the equation holds true as can be shown for (2) and (3).
19. Three lattice planes for the simple cubic lattice are depicted in Fig
8.15. Of these, ﬁgure 8.15(b) represents a member of the family with Miller
indices (110). Therefore, the plane is perpendicular to the reciprocal lattice
vector:
g = b
1
+b
2
Equation 8.33 gives the distance between planes:
d =
2π
|g|
Copyright © 2010 by Elsevier, Inc. All rights reserved.
61
We may ﬁnd the magnitude of g using the primitive reciprocal vectors found
in example 8.1:
g =
2π
a
ˆ
i +
2π
a
ˆ
j ⇒|g| =
√
2
2π
a
Therefore our result:
d =
a
√
2
=
√
2a
2
21. a) First assume g = g

:
_
a
0
e
−igx
e
igx
dx =
_
a
0
dx = a
On the other hand, if g = g

our integral is as follows:
_
a
0
e
−ig

x
e
igx
dx =
_
a
0
e
ix(g−g

)
dx
By using Euler formula, the integrand may be expressed as a combination of
sine and cosine functions:
_
a
0
cos(x(g −g

)) +i sin(x(g −g

))dx
Therefore, from equation 8.10 we see that the arguments of the cos and sin
functions will be of the form:
2πn
a
x
Therefore, the integral from 0 to a is over a full period. Such integrals of cos
and sin functions are equal to zero.
b) Eq. 8.14 gives:
f(x) =

g
F
g
e
igx
Multiply by e
−ig

x
:
e
−ig

x
f(x) =

g
F
g
e
−ig

x
e
igx
Copyright © 2010 by Elsevier, Inc. All rights reserved.
62 8. ELECTRONIC STRUCTURE OF SOLIDS - SOLUTIONS
Integrate from 0 to a:
_
a
0
e
−ig

x
f(x) dx =

g
F
g
_
a
0
e
−ig

x
e
igx
dx
From part (a), we know that the integral on the right-hand side has a value
of zero unless g = g

, in which case the value of the integral is ”a”. Thus,
the summation disappears as the only term that survives is g = g

:
_
a
0
e
−ig

x
f(x) dx = aF
g

25. We shall ﬁrst carry out the substitutions then apply orthogonality
conditions as directed. Our terms as given in the text are:
Eq. 8.55 : ψ
k
(r) =
1
V
1/2
e
ik·r
Eq. 8.58 : V (r) =

g
V
g
e
ig·r
Our other two needed quantities may be determined from these:
ψ

k
(r) =
1
V
1/2
e
ik

·r
and ψ
∗
k
(r) =
1
V
1/2
e
−ik·r
Substituting into the integral:
I
∗
k

, k
=

g
V
g
V
_
e
i(g+k

−k)·r
dV
We separate the exponential terms, so the the integral may be written:
I
∗
k

, k
=

g
V
g
V
_
e
i(g+k

)·r
e
−ik·r
dV
We may also think of this as:

g
V
g
V
_
ψ
∗
k
(r)ψ
(k

+g)
(r) dV
Using the orthogonality condition of 8.52:
k = k

+g

which is equation 8.62.
Copyright © 2010 by Elsevier, Inc. All rights reserved.
10
Semiconductor Lasers -
Solutions
1. As with other radiative transitions we have dealt with, the photons emit-
ted must have the same energy as the diﬀerence in energies between the two
states. Here, the two states are the bottom of the conduction band and the
top of the valence band, with an energy diﬀerence ∆ = 1.519 eV . Using
equation I.25:
E =
hc
λ
λ =
hc
E
=
1240 eV · nm
1.519 eV
= 816.3nm
3. As outlined in section 10.3.1, heterostructures may be formed using semi-
conductors with similar lattice constants. Thus, the heterostructures grown
on a substrate of InP must have the same lattice constant as InP. We’ll use
the linear interpolation procedure for each alloy:
a(In
x
Ga
1−x
As) = a(InP) = xa(InAs) + (1 −x)a(GaAs)
5.869 = 6.058x + 5.653(1 −x) ⇒x = .53
This leads to an ally of composition: In
.53
Al
.48
As
5. By examining ﬁgure 2.6, we see that the graph of
_
(θ
2
0
/θ
2
) −1 inter-
sects the graph of tan θ and −cot θ three times as it decays. Therefore, if
63
Copyright © 2010 by Elsevier, Inc. All rights reserved.
64 10. SEMICONDUCTOR LASERS - SOLUTIONS
we insure that
_
(θ
2
0
/θ
2
) −1 reaches zero after it intersects once, but before
it does again, the well will have only one bound state. Per equation 2.30, it
θ
2
0
that includes dependence on V
0
, the depth of the well. So, we solve for θ
0
where the whole expression equals zero:
_
θ
2
0
θ
2
−1 = 0 ⇒θ
2
0
= θ
2
We kepp only the positive solutions and impose the condition that θ < π/2
as that is where the graph of −cot θ begins in the positive domain. If
_
(θ
2
0
/θ
2
) −1 decays completely before this point, then the electron will not
have more than one ground state.
θ <
π
2
⇒θ
2
0
<
_
π
2
_
2
We now solve eq. 2.30 for V
0
, using SI units at ﬁrst:
θ
2
0
=
m
0
V
0
L
2
2
2
<
π
2
⇒V
0
<
π
2
m
0
L
2
π
2
m
0
L
2
=
(3.1416)(1.055 ×10
−34
J · s)
2
.067(9.109 ×10
−31
kg)(10 ×10
−9
m)
2
= 5.729 ×10
−21
J
We should now convert to eV:
5.729 ×10
−21
J ·
1 eV
1.6 ×10
−19
eV/J
= .036 eV
The well must be less than 0.036 eV deep to insure that no more than one
bound state exists.
7. We will use the following procedure for these matrix multiplication oper-
ations:
_
_
A
1
A
2
A
3
B
1
B
2
B
3
C
1
C
2
C
3
_
_
_
_
D
1
D
2
D
3
_
_
=
_
_
A
1
D
1
+A
2
D
2
+A
3
D
3
B
1
D
1
+B
2
D
2
+B
3
D
3
C
1
D
1
+C
2
D
2
+C
3
D
3
_
_
_
_
1 2 0
1 1 2
1 3 1
_
_
_
_
1
1
0
_
_
=
_
_
3
2
4
_
_
Copyright © 2010 by Elsevier, Inc. All rights reserved.
65
_
_
1 0 1
1 2 1
1 1 3
_
_
_
_
1
0
1
_
_
=
_
_
2
2
4
_
_
_
_
2 1 1
1 0 1
1 1 0
_
_
_
_
1
2
1
_
_
=
_
_
5
2
3
_
_
9. We shall proceed left to right, ﬁrst consolidating the 1/2k
1
and 1/2k
2
terms as we will reintroduce them after carrying out the matrix operations.
_
k
1
+k
2
k
1
−k
2
k
1
−k
2
k
1
+k
2
_ _
e
ik
2
L
0
0 e
−ik
2
L
_
=
_
e
ik
2
L
(k
1
+k
2
) e
−ik
2
L
(k
1
−k
2
)
e
ik
2
L
(k
1
−k
2
) e
−ik
2
L
(k
1
+k
2
)
_
Using this result in eq. 10.30:
T =
1
4k
1
k
2
_
e
ik
2
L
(k
1
+k
2
) e
−ik
2
L
(k
1
−k
2
)
e
ik
2
L
(k
1
−k
2
) e
−ik
2
L
(k
1
+k
2
)
_ _
k
1
+k
2
k
2
−k
1
k
2
−k
1
k
1
+k
2
_
Carrying this through will give us following:
T =
_
T
11
T
12
T
21
T
22
_
where each element is as follows:
T
11
=
1
4k
1
k
2
_
e
ik
2
L
(k
1
+k
2
)
2
−e
−ik
2
L
(k
1
−k
2
)
2
¸
T
12
=
−1
4k
1
k
2
_
e
ik
2
L
(k
2
1
−k
2
2
)
2
−e
−ik
2
L
(k
2
1
−k
2
2
)
¸
T
21
=
1
4k
1
k
2
_
e
ik
2
L
(k
2
1
−k
2
2
) −e
−ik
2
L
(k
2
1
−k
2
2
)
¸
T
22
=
−1
4k
1
k
2
_
e
ik
2
L
(k
1
−k
2
)
2
−e
−ik
2
L
(k
1
+k
2
)
2
¸
We see that the expression we found for T
11
does in fact agree with the ﬁrst
part of eq. 10.31, however the Euler formula was used in the second part to
Copyright © 2010 by Elsevier, Inc. All rights reserved.
66 10. SEMICONDUCTOR LASERS - SOLUTIONS
further simplify:
T
11
=
1
4k
1
k
2
_
e
ik
2
L
(k
1
+k
2
)
2
−e
−ik
2
L
(k
1
−k
2
)
2
¸
=
1
4k
1
k
2
_
(cos k
2
L +i sin k
2
L)(k
2
1
+ 2k
1
k
2
+k
2
2
) −(cos k
2
L −i sin k
2
L)(k
2
1
−2k
1
k
2
+k
2
2
)
¸
=
1
2k
1
k
2
[2k
1
k
2
cos k
2
L +i(k
2
1
+k
2
2
)sin k
2
L]
The same may be done for the remaining three elements if desired.
11. Equations10.25 and 10.26 give the following:
A
1
= T
11
A
2
+T
12
B
2
B
1
= T
21
A
2
+T
22
B
2
The transmission amplitude of light incident from the right will be equal to
B
1
/B
2
and the corresponding reﬂection amplitude is equal to A
2
/B
2
. As
directed, we set A
1
= 0.
0 = A
1
= T
11
A
2
+T
12
B
2
(1)
B
1
= T
21
A
2
+T
22
B
2
(2)
Solving (1) for A
2
and substituting into (2):
B
1
=
−T
21
T
12
B
2
T
11
+T
22
B
2
= B
2
1
T
11
(T
11
T
22
−T
21
T
12
)
We recognize the quantity in parentheses as the determinant of the T-matrix,
therefore:
t
21
=
B
1
B
2
=
det T
T
11
This equation 10.38. Now for 10.39, we solve eq. (1) above for A
2
/B
2
:
r
21
=
A
2
B
2
=
−T
12
T
11
15. a) The total rate of change in the amount of water in the reservoir
Copyright © 2010 by Elsevier, Inc. All rights reserved.
67
will equal the rate of water entering minus the rate in which it is leaving.
Letting V equal the volume of water in the reservoir:
dV
dt
= R
f
−R
D
Where R
D
is the total rate at which water drains. Since the reservoir is
draining from two ports, we set up the following:
R
D
= R
d1
+R
d2
Therefore, our total rate equation is:
dV
dt
= R
f
−h(C
1
+C
2
)
b) To ﬁnd the steady state height, we set:
dV
dt
= 0
And solve for h:
h =
R
f
c
1
+c
2
Copyright © 2010 by Elsevier, Inc. All rights reserved.
68 10. SEMICONDUCTOR LASERS - SOLUTIONS
Copyright © 2010 by Elsevier, Inc. All rights reserved.
11
Relativity I - Solutions
1. Einstein’s second postulate on the special theory of relativity states that
the speed of light is independent of the motion of the source, therefore the
velocity of the light in the reference frame of the spaceship is c.
3. a) The inverse Lorentz transformations are given by eqs. 11.17. There-
fore, we need to ﬁrst calculate γ:
γ =
1
1
√
1−u
2
/c
2
=
1
√
1 −.25
2
= 1.033
Now we transform from S to S

:
x = γ(x

+ut

) = 1.033 (2m+.25(4)m)
= 3.10 m
y = y

= 0 m
z = z

= 0 m
t = γ(t

+ux

/c
2
) =
1.033
c
(4 s +.5 s)
=
4.65 s
c
b) We now use eqs. 11.15 and the results from part (a) to transform back
69
Copyright © 2010 by Elsevier, Inc. All rights reserved.
70 11. RELATIVITY I - SOLUTIONS
to S

:
x

= γ(x −ut) = 1.033(3.10 m−.25(4.65) m)
= 2.0 m
y

= y = 0 m
z

= z = 0 m
t

= γ(t −ux/c
2
) =
1.033
c
(4.39 s −.25(3.10) s)
=
4.0 s
c
5. a) We may use equation 11.23 to calculate the length as perceived in
our reference frame:
L
M
=
_
1 −
.5c
c
2
(1 m) = .86 m
b) Equation 11.26 will give us the time we record the object passing us.
However, we need to calculate ∆t
R
, which is the time we would record if we
were traveling along with meter stick:
∆t
R
=
d
v
=
11 m
.5(3 ×10
8
m/s)
= 6.67 ×10
−9
s
Now, to transform to our frame of reference:
∆t
M
=
6.67 ×10
−9
s
_
1 −
(.5c)
2
c
2
= 7.70 ×10
−9
s
7. The percentage of length contraction will be equal to:
_
1 −
L
M
L
R
_
×100%
Therefore, we may calculate L
M
/L
R
via eq. 11.23:
L
M
L
R
=
_
1 −
u
2
c
2
=
_
1 −
(3 ×10
−6
c)
2
c
2
Copyright © 2010 by Elsevier, Inc. All rights reserved.
71
L
M
L
R
≈ 1 ⇒0% contraction
While the high speed of the jet is truly impressive, they still do not travel with
enough velocity to display relativistic eﬀects. The time dilation experienced
here may be calculated by equation 11.26:
∆t
M
=
∆t
R
1
√
1−u
2
/c
2
=
1 yr
_
1 −
(3×10
−6
c)
2
c
2
∆t
M
≈ 1 yr
Once again, the relativistic time dilation has an extremely small eﬀect, well
within the intrinsic error of any wristwatch.
9. a) By diﬀerentiating both sides of the equation given in the problem
for the position of the center of mass with respect to t, we may calculate the
velocity of the center of mass:
d
dt
2mx
CM
=
d
dt
(mx
e
+mx
p
)
dx
CM
dt
=
1
2
(u
e
+u
p
)
where u
e
and u
p
equal the velocities of the electtron and positron, respec-
tively. We now substitute these quantities to obtain the velocity of the center
of mass:
u
x
=
1
2
(.95c +.2c) = .575c
b) The lifetime of the produced particle, gives 2.0 × 10
−8
s will be taken
as ∆t
R
in eq. 11.26:
∆t
M
=
2.0 ×10
−8
s
_
1 −
(.575c)
2
c
2
= 2.44 ×10
−8
s
11. a) The equations of section 11.4.6 will give us the transformed frequen-
cies, however we are given the wavelength of the sodium D
2
line. This is a
minor obstacle, as we can use equation I.23:
f
0
=
c
λ
0
=
3 ×10
8
m/s
589.0 ×10
−9
m
= 5.09 ×10
14
Hz
Copyright © 2010 by Elsevier, Inc. All rights reserved.
72 11. RELATIVITY I - SOLUTIONS
For an approaching light source:
f =
¸
1 +β
1 −β
f
0
where β = v/c = 0.3.
f =
_
1.3
0.7
(5.09 ×10
14
Hz) = 6.94 ×10
14
Hz
We then use eq. I.23 to calculate the correspond wavelength:
λ =
c
f
=
3 ×10
8
m/s
6.94 ×10
14
Hz
= 432.2 nm
λ −λ
0
= 432.2 nm−589.0 nm = −156.8 nm
b) For a receding light source:
f =
¸
1 −β
1 +β
f
0
=
_
.7
1.3
(5.09 ×10
14
Hz) = 3.74 ×10
14
Hz
λ =
c
f
=
3 ×10
8
m/s
3.74 ×10
14
Hz
= 802.1 nm
λ −λ
0
= 802.1 nm−589.0 nm = 213.1 nm
c) The equation adjacent to ﬁgure 11.12 allows us to calculate the Doppler
shift due to transverse motion:
f =
f
0
γ
We calculate γ:
γ =
1
_
1 −u
2
/c
2
=
1
_
1 −(.3)
2
= 1.05
And substitute:
f =
5.09 ×10
14
Hz
1.05
= 4.86 ×10
14
Hz
Copyright © 2010 by Elsevier, Inc. All rights reserved.
73
λ =
c
f
=
3 ×10
8
m/s
4.86 ×10
14
Hz
= 617.2 nm
λ −λ
0
= 617.2 nm−589.0 nm = 28.2 nm
13. We use eq. 11.41 for the receding source:
β =
v
c
= 0.80
f =
_
1 −.80
1 +.80
×100 MHz = 33 MHz
15. Equation 11.16 gives the deﬁnition for γ:
γ =
1
_
1 −u
2
/c
2
However, we ﬁrst solve eq. 11.38 for v:
v

= γ(v −u) · γ(1 +uv

/c
2
)
We ﬁrst simplify the right-hand side, then bring the terms containing v

to
the left, and solve:
v

= γ
2
_
v −u +
vuv

c
2
−
u
2
v

c
2
_
v

γ
2
−
v

vu
c
2
+
u
2
v

c
2
= v −u
v

_
1
γ
2
−
uv
c
2
+
u
2
c
2
_
= v −u
v

_
1 −
u
2
c
2
−
uv
c
2
+
u
2
c
2
_
= v −u
v

=
v −u
1 −
uv
c
2
Copyright © 2010 by Elsevier, Inc. All rights reserved.
74 11. RELATIVITY I - SOLUTIONS
17. Eq. 11.43 allows us to calculate the proper time interval. We apply
this equation at each segment of the observer’s journey. First, the traveler
remains stationary for 1 m of time:
(proper time)
2
= 1
2
−0
2
proper time = 1 m
Then, he travels 4 m in 5 m of time:
(proper time)
2
= 5
2
−4
2
proper time = 3 m
Finally he travels 1 m in 2 m of time:
(proper time)
2
= 2
2
−1
2
proper time =
√
3 m
Therefore, the total proper time is 4 +
√
3 m.
22. Per equation 11.60, t
σ
= g
σµ
t
µ
, and eq. 11.58, t
µ
= g
µν
t
ν
, the g matrices
will serve to raise or lower the indices of a vector. We now use the given
expression Λ
ν
µ
= g
µρ
Λ
ρ
σ
g
σν
to operate on v
ν
:
g
σν
v
ν
= v
σ
Now, the Λ
ρ
σ
will serve to transform to the ρ

indices:
Λ
ρ
σ
g
σν
v
ν
= Λ
ρ
σ
v
σ
= v
ρ
We now use g
µρ
to lower the index:
g
µρ
Λ
ρ
σ
g
σν
v
ν
= g
µρ
v
ρ
= v

ρ
We have now shown that v

µ
= Λ
ν
µ
v
ν
, which is similar to the contravariant
vector’s transformation in form but requires an alternately deﬁned transfor-
mation matrix Λ
ν
µ
.
Copyright © 2010 by Elsevier, Inc. All rights reserved.
75
23. First recall that g
µρ
and g
σν
will have the form:
_
¸
¸
_
1 0 0 0
0 −1 0 0
0 0 −1 0
0 0 0 −1
_
¸
¸
_
The ﬁrst operation, g
µρ
Λ
ρ
σ
:
_
¸
¸
_
1 0 0 0
0 −1 0 0
0 0 −1 0
0 0 0 −1
_
¸
¸
_
_
¸
¸
_
γ −βγ 0 0
−βγ γ 0 0
0 0 1 0
0 0 0 1
_
¸
¸
_
=
_
¸
¸
_
γ −βγ 0 0
βγ −γ 0 0
0 0 −1 0
0 0 0 −1
_
¸
¸
_
Now, g
µρ
Λ
ρ
σ
g
σν
:
_
¸
¸
_
γ −βγ 0 0
βγ −γ 0 0
0 0 −1 0
0 0 0 −1
_
¸
¸
_
_
¸
¸
_
1 0 0 0
0 −1 0 0
0 0 −1 0
0 0 0 −1
_
¸
¸
_
=
_
¸
¸
_
γ βγ 0 0
βγ γ 0 0
0 0 1 0
0 0 0 1
_
¸
¸
_
Copyright © 2010 by Elsevier, Inc. All rights reserved.
76 11. RELATIVITY I - SOLUTIONS
Copyright © 2010 by Elsevier, Inc. All rights reserved.
12
Relativity II - Solutions
1. The components of the four-velocity are given by the relations between
eqs. 12.3 and 12.4. Since the electron is traveling in the x-direction, we note
that:
dy
dt
=
dz
dt
= 0
Therefore, our relevant components are:
v
0
= γc and v
1
= γ
dx
dt
We then calculate γ:
γ =
1
_
1 −u
2
/c
2
=
1
_
1 −
(.2c)
2
c
2
= 1.02
Giving us:
v
0
= 1.02c
v
1
= 1.02(.2c) = .204c
b. The four-momentum as given in eq. 12.6 is simply the product of the
electron’s mass and each component of its four-velocity:
p
0
= 1.02m
e
c
p
1
= .204m
e
c
77
Copyright © 2010 by Elsevier, Inc. All rights reserved.
78 12. RELATIVITY II - SOLUTIONS
3. We may use equations 12.12 and 12.13 which give expressions for the
rest energy and kinetic energy, respectively. Setting R=KE:
γmc
2
= 2(γ −1)mc
2
γ = 2γ −2 ⇒γ = 2
Using eq. 12.3 for γ:
γ =
1
_
1 −u
2
/c
2
= 2
_
(1 −v
2
/c
2
) =
1
2
_
1 −
v
2
c
2
_
=
1
4
v
2
c
2
=
3
4
v =
√
3
2
c
In order for the kinetic evergy to be equal to the rest energy, the particle
must be traveling at a speed of (
√
3/2)c.
5. We ﬁrst evaluate the derivatives of f(x) called for in the Taylor series:
f

(x) =
1
2
1
(1 −x)
3/2
f

(x) =
3
4
1
(1 −x)
5/2
We now evaluate at x=0 and substitute into the given form of the Taylor
series:
f(x) = 1 +
1
2
x =
3
8
x
2
+. . .
Now, let x = v
2
/c
2
:
1
_
1 −u
2
/c
2
= 1 +
1
2
v
2
c
2
=
3
8
v
4
c
4
Copyright © 2010 by Elsevier, Inc. All rights reserved.
79
This result is identical to eq. 12.10.
7. We shall use eq. 12.13 and let m equal the electron’s mass. However,
it will be most convenient to use the mass given in Appendix A in units of
MeV/c
2
.
100 MeV = (γ −1)
_
.511
MeV
c
2
_
c
2
100 = .511γ −.511
γ = 196.7
We now ﬁnd the electron’s speed from eq. 12.3:
γ = 196.7 =
1
_
1 −u
2
/c
2
38688 =
1
1 −
v
2
c
2
1 −
v
2
c
2
= 2.58 ×10
−5
.99997 =
v
2
c
2
v ≈ .99999c
The electron would have to be traveling toughly the spped of light possess
100 MeV of kinetic energy.
9. a) For the laboratory frame, recall that the target particle is at rest,
therefore has zero momentum. We go ahead and carry out the four-vector
addition as called for in the numerator of the expression for s:
|p
A
+p
B
|
2
c
2
=
[(E
A
+E
B
)
2
−(p
A
+p
B
)
2
]
c
2
Note that we have separated the energy and momentum components of the
four-momentum. Since particle B is at rest, we realize that E
B
must equal
the rest energy of the particle and p
B
= 0:
s =
[(E
A
+m
B
c
2
)
2
−|p
A
|
2
]
c
2
Copyright © 2010 by Elsevier, Inc. All rights reserved.
80 12. RELATIVITY II - SOLUTIONS
In the center of mass frame, we take the colliding particles to have equal and
opposite momenta. Therefore, we begin as we did for the laboratory frame,
however we can no longer take the energy of particle B to be equal to its rest
energy and we allow the sum of the momenta to be zero.
s =
[(E
A
+E
B
)
2
−(p
A
+p
B
)
2
]
c
2
=
(E
A
+E
B
)
2
c
2
b) We begin by directly substituting the sums of the four-vectors into the
sum of the Mandelstam variables:
s +t +u =
1
c
2
_
(E
A
+E
B
)
2
c
2
−(p
A
+p
B
)
2
+
(E
A
−E
C
)
2
c
2
−(p
A
−p
C
)
2
+
(E
A
−E
D
)
2
c
2
−(p
A
−p
D
)
2
_
By conservation of energy and momentum:
E
A
−E
D
= E
C
−E
B
E
A
−E
C
= E
D
−E
B
p
A
−p
D
= p
C
−p
B
p
A
−p
C
= p
D
−p
B
We substitute these quantities into our equation above:
s +t +u =
1
c
2
_
(E
A
+E
B
)
2
c
2
−(p
A
+p
B
)
2
+
(E
D
−E
B
)
2
c
2
−(p
D
−p
B
)
2
+
(E
C
−E
B
)
2
c
2
−(p
C
−p
B
)
2
_
We use the fact that in the laboratory frame, p
B
= 0 and E
B
= m
B
c
2
and
simplify:
s +t +u =
2E
A
m
B
c
2
−
2E
D
m
B
c
2
−
2E
C
m
B
c
2
+m
2
A
+ 3m
2
B
+m
2
C
+m
2
D
=
2m
B
c
2
(E
A
−E
D
−E
C
) +m
2
A
+ 3m
2
B
+m
2
C
+m
2
D
Copyright © 2010 by Elsevier, Inc. All rights reserved.
81
Once again, from the conservation of energy, we know that E
B
= E
A
−E
D
−
E
C
. Making this substitution and using E
B
= m
B
c
2
:
s +t +u = −2m
2
B
+m
2
A
+ 3m
2
B
+m
2
C
+m
2
D
= m
2
A
+m
2
B
+m
2
C
+m
2
D
11. The particle collision here involes two particles with equal mass and
speed traveling in the opposite direction. Therefore, using the conservation
of momentum:
p
A
+p
B
= p
C
= 0
Thus, we are in the center of mass frame. An important result here is that
the resulting particle is at rest. We may then use the conservation energy
where the energy of the resulting particle is its rest energy and the energy
of the colliding particles are given by eq. 12.9. We let m be the mass of the
each of the colliding particles and m
C
be the mass of the resulting particle:
E
A
+E
B
= E
C
⇒2γmc
2
= m
C
c
2
m
C
= 2γm
Using equation 12.3:
m
C
= 2
m
_
1 −(2/3)
2
= 2.68m
We see that since the mass of the resulting particle is greater than the sum
of the colliding particles, some of their kinetic energy was converted into the
resulting particle’s rest-energy.
13. We consider the center of mass frame where the proton and antiproton
have equal and opposite momenta. Since the proton and antiproton have
equal masses, their speeds must also be equal. Therefore, when we impose
conservation of energy and use eq. 12.9:
E
p
+E
¯ p
= E
X
where E
X
is the energy of the resulting particle. As stated above, we are
in the center of mass frame, where the resulting particle is at rest and its
energy is given by eq. 12.12:
2γm
p
c
2
= m
X
c
2
Copyright © 2010 by Elsevier, Inc. All rights reserved.
82 12. RELATIVITY II - SOLUTIONS
2γ(938 MeV ) = 9700 MeV
γ = 5.17
We may then ﬁnd the kinetic energies of the colliding particles by equation
12.13:
KE = (γ −1)m
p
c
2
= 3912 MeV
Their velocities may be calculated by eq. 12.3:
γ = 5.17 =
1
_
1 −u
2
/c
2
1 −
v
2
c
2
= .0374
v = .981c
The proton and antiproton will each have a speed of .981c in the center of
mass frame.
15. First, we test the relation:
α
i
α
j
+α
j
α
i
= 2δ
ij
I
The Dirac-Pauli representation is given in eq. 12.42 and we merely substitute:
α
i
α
j
=
_
0 σ
i
σ
i
0
_ _
0 σ
j
σ
j
0
_
=
_
σ
i
σ
j
0
0 σ
i
σ
j
_
α
j
α
i
=
_
0 σ
j
σ
j
0
_ _
0 σ
i
σ
i
0
_
=
_
σ
j
σ
i
0
0 σ
j
σ
i
_
α
i
α
j
+α
j
α
i
=
_
σ
i
σ
j
+σ
j
σ
i
0
0 σ
i
σ
j
+σ
j
σ
i
_
= (σ
i
σ
j
+σ
j
σ
i
)I
It can be shown that the Pauli matrices satisfy the following condition:
σ
i
σ
j
=
_
I if i = j
−σ
j
σ
i
if i = j
Therefore, we see that:
σ
i
σ
j
+σ
j
σ
i
=
_
2 if i = j
0 if i = j
Copyright © 2010 by Elsevier, Inc. All rights reserved.
83
Or, more concisely, σ
i
σ
j
+σ
j
σ
i
= 2δ
ij
I. Thus, α
i
α
j
+α
j
α
i
= 2δ
ij
I .
Now, we turn our attention to the relation: β
2
= I. From eq. 12.42, we
get the deﬁnition for β:
β =
_
I 0
0 I
_
We carry out the multiplication:
_
I 0
0 I
_ _
I 0
0 I
_
=
_
I
2
0
0 I
2
_
Since I
2
= I, β
2
= I.
17. Muons and electrons are both Dirac particles. The Feynman diagrams
for muon-electron scattering are equivalent to the Feynman diagram shown
in Fig. 12.4 with one of the incoming and outgoing lines corresponding to
an electron and the other corresponding to a muon.
Copyright © 2010 by Elsevier, Inc. All rights reserved.
84 12. RELATIVITY II - SOLUTIONS
Copyright © 2010 by Elsevier, Inc. All rights reserved.
13
Particle Physics - Solutions
1. a) This process can not occur as it would violate the conservation of
baryon number and lepton number. The baryon number is one of the left-
hand side but zero on the right. Likewise, the electron lepton number of the
left-hand side is zero while L
e
= −1 on the right.
b) This process can not occur due to violation of energy conservation. The
Σ
◦
has a rest energy of 1192 MeV while the Λ and the π
◦
have a total rest
energy of 1250 MeV.
c) Conservations of charge, energy, baryon number, and lepton number are
all observed, therefore the process may occur. Furthermore, the process con-
serves strangeness and proceeds by the strong interaction.
d) Here, the conservation of muon lepton number is violated as the total
number on the left is 0, but is -1 on the right.
e) K
−
has a strange quantum number S=-1 where the π
−
π
◦
system has
S=0. Since charge, energy, lepton number, and baryon number are all con-
served, this process proceeds via the weak interaction.
f ) The ρ
0
, being a boson, will have a symmetric wave function. The spin of a
ρ
0
will have a value of 1, therefore its total angular momentum will be J=1.
By the conservation of angular momentum, J=1 for the right-hand side, thus
the spinless pions would have an antisymmetric total wave function. Also
being bosons, this is forbidden.
85
Copyright © 2010 by Elsevier, Inc. All rights reserved.
86 13. PARTICLE PHYSICS - SOLUTIONS
g) Strangeness is not conserved while the other conservation laws are, there-
fore the process must occur via the weak interaction.
3. The rest energies may be found in tables 13.9 through 13.11.
a) Initial state: 140 + 938 = 1078 MeV
Final state: 940 + 135 = 1074 Mev
b) Initial state: 140 + 938 = 1116 MeV
Final state: 938 + 135 = 1073 Mev
c) Initial state: 140 + 938 = 1078 MeV
Final state: 494 + 1138 = 1632 Mev
The kinetic energy of this initial state must be greater than that of the
ﬁnal state to ensure conservation of energy.
5 To calculate the strangeness of these particles, we refer to tables 13.9-
13.11 to ﬁnd the quark content of each particle, then use eq. 13.26.
a) The Ω
−
has a quark content of sss, therefore using equation 13.26:
S = −[3] = −3
We repeat for the particles on the right-hand side to ﬁnd that S(Λ) = −1
and S(K
−
) = −1. Therefore ∆S = 1 and the process must occur via the
weak interaction.
b) Using the same procedure as part (a), S(Σ
◦
) = −1, S(Λ) = −1, and
S(γ) = 0. There is no change in strangeness, therefore the process may oc-
cur via the electromagnetic, as we recognize the presence of the photon.
c) Using the same procedure as parts (a) and (b) above, S(Λ) = −1,
S(p) = 0, and S(π
−
)=0. ∆S = 1 and the process must occur via the weak
interaction.
Copyright © 2010 by Elsevier, Inc. All rights reserved.
87
7. a) This process may occur via the strong interaction as all conserva-
tion laws are met.
b) The strangeness of the initial state is zero; however, the strangeness has a
value of -1 in the ﬁnal state. The other conservation laws are observed, and
therefore this process must occur via the weak interaction.
c) This process is strictly forbidden as the baryon number of the initial
state is +1 while that of the ﬁnal is zero. The Λ particle has baryon number
of +1, while the
¯
Σ
◦
antiparticle has a baryon number of -1.
d) This process is strictly forbiden as the muon lepton number of the an-
tineutrino is -1 which is the total muon lepton number for the initial state.
The ﬁnal state has a muon lepton number zero. Note that the similar con-
servation of electron lepton number is also broken.
e) Here again, lepton number conservation is violated. The electron neu-
trino on the left-hand side has an electron lepton number of +1 whereas the
positron on the right-hand side has an electron lepton number of -1. As a
consequence, this process is strictly forbidden.
9. The quark composition may be found in tables 13.9-13.11:
a)
Ω
−
: sss Λ : uds K
−
: s¯ u
Here we have nothing but strange quarks in the initial state, but the ﬁnal
state contains one fewer s quark but does contain u, ¯ u, d quarks.
b)
π
−
: dˆ u p : uud Λ : uds K
◦
: s
¯
d
There are no strange quarks in the inital state, but two in the ﬁnal. Notice
that the antiquark in the left-hand meson is a ¯ u, while the antiquark that
appears in the meson that occurs on the right is an antidown quark.
c)
p : uud K
−
: s¯ u Ξ
−
: dss K
+
: u¯ s
Copyright © 2010 by Elsevier, Inc. All rights reserved.
88 13. PARTICLE PHYSICS - SOLUTIONS
There is no change in strangeness during this process. Even though there
is an additional strange quark on the right, there is also an ¯ s quark which
cancels the added strange quantum number.
11. If the wave has periodic boundary conditions at the two ends of the
region, then the length L must equal an integer multiple of the wavelength:
L = nλ ⇒λ =
L
n
Using the de Broglie relation and this value of the wavelength:
p =
h
λ
=
hn
L
Recall that = h/2π therefore:
p =
2πn
L
We can think of this relation in the following way. In momentum space, each
state has a “length” of 2π/L, therefore, the number of states in a diﬀerential
length dp will be:
number of states =
dp
2π/L
=
L
2π
dp (1)
If we now expand our region to include a cubic volume, extending to x = L,
y = L, z = L from the origin, we now have three terms like (1) that form a
volume element in a momentum space:
dN =
L
2π
dp
x
L
2π
dp
y
L
2π
dp
x
=
V
(2π)
3
d
3
p
13. We can use ﬁgure 13.29 as a guide to obtain the remaining weights
from the highest weight. Notice in ﬁgure 13.29(b) the highest weight is 3µ
1
,
where µ
1
is given by eq. 13.84. As we work along a diagonal such as the one
with the weights labeled in Fig. 13.29(b), we subtract an additional α
1
. As
Copyright © 2010 by Elsevier, Inc. All rights reserved.
89
we work right-to-left along a row, we subtract (α
1
+ α
2
) for each step. As
follows, we shall work from right to left across a row, and completing rows
from the top down to 3µ
1
−3α
1
. Expressions for α
1
and α
2
are given in eq.
13.78:
3µ
1
= 3
_
1
√
3/3
_
=
_
3
√
3
_
3µ
1
−α
1
−α
2
=
_
2
√
3
_
3µ
1
−2α
1
−2α
2
=
_
1
√
3
_
3µ
1
−3α
1
−3α
2
=
_
0
√
3
_
3µ
1
−α
1
=
_
5/2
√
3/2
_
3µ
1
−2α
1
−α
2
=
_
3/2
√
3/2
_
3µ
1
−3α
1
−2α
2
=
_
5/2
√
3/2
_
3µ
1
−2α
1
=
_
2
0
_
3µ
1
−3α
1
−α
2
=
_
1
0
_
3µ
1
−3α
1
=
_
3/2
−
√
3/2
_
Copyright © 2010 by Elsevier, Inc. All rights reserved.
90 13. PARTICLE PHYSICS - SOLUTIONS
Copyright © 2010 by Elsevier, Inc. All rights reserved.
14
Nuclear Physics - Solutions
1. a) Recall from section 14.2 the nomenclature for various isotopes:
A
Z
X
where A atomic mass number which is the sum of the number of neutrons
and number of protons, while z is the atomic number which is the number of
protons, and X is the element symbol. Therefore, the number of neutrons is
equal to A minus Z:
3
7
Li : 3 protons, 4 neutrons
63
29
Cu : 29 protons, 34 neutrons
238
92
U : 92 protons, 146 neutrons
3. Equation 14.3 provides an estimate of the radius of a nucleus:
R = 1.12A
1/3
fm
where A is the atomic mass number:
4
2
He : R = 1.12(4)
1/3
fm = 1.78fm
16
8
He : R = 1.12(16)
1/3
fm = 2.82fm
56
26
Fe : R = 1.12(56)
1/3
fm = 4.28fm
208
82
Fe : R = 1.12(208)
1/3
fm = 6.64fm
237
93
Fe : R = 1.12(237)
1/3
fm = 6.93fm
91
Copyright © 2010 by Elsevier, Inc. All rights reserved.
92 14. NUCLEAR PHYSICS - SOLUTIONS
8. The semi-empirical formula without the pairing term is as follows:
B(N, Z) = aA −bA
2/3
−
dZ
2
A
1/3
−s
(N −Z)
2
A
If N=Z, we see that the fourth term immediately equals zero. Since we want
the binding energy per nucleon, we can divide the entire equation by the
number of nucleons, which will be equal to A:
B(N, Z)
A
= a −ba
−1/3
−
dZ
2
A
4/3
Now we let Z=A/2:
Binding energy per nucleon = a −bA
−1/3
−
dA
2
4A
4/3
= a −bA
−1/3
−
d
4
A
2/3
Now, we diﬀerentiate with respect to A and set equal to zero:
d
¯
B
dA
=
1
3
A
−4/3
−
1
2
dA
−1/3
= 0
Now solve for A:
1
3
A
−1/3
(bA
−1
−
2
d
) = 0
b
A
=
2
d
A =
2b
d
The critical points are A = 0 and A = 51.3. Per the veriﬁcation called
for in the problem, A/2 ≈ 26. Further testing of the value of the derivative
in the intervals (0, 13) and (13, ∞) further identiﬁes A/2 = 26 as a maximum.
9. We use equation 14.13 ﬁrst which relates the half-life to the propor-
tionality constant λ:
t
1/2
= 2 min = 120 s =
ln2
λ
λ =
ln2
120 s
= .006 s
−1
Copyright © 2010 by Elsevier, Inc. All rights reserved.
93
We now use the equation following eq. 14.12 and our initial condition R =
1200 s
−1
at t=0 to ﬁnd N
0
:
R = 1200s
−1
= .006N
0
e
−.006(0)
= (.006s
−1
)N
0
N
0
= 2 ×10
5
Now we may use this same equation to ﬁnd the transition rates at the sub-
sequent times:
t = 4 min = 240 s : R = (.006s
−1
)(2 ×10
5
)e
−.006(240)
R = 284 s
−1
t = 6 min = 360 s : R = (.006s
−1
)(2 ×10
5
)e
−.006(360)
R = 138 s
−1
t = 8 min = 480 s : R = (.006s
−1
)(2 ×10
5
)e
−.006(480)
R = 67 s
−1
We see, just as expected, that the decay rate itself decays exponentially.
11. Our strategy here should be to ultimately use the rate of decay equation
(the one immediately following eq. 14.12) to ﬁnd the number of
14
6
C atoms
that are responsible for the measured radiation in the sample. We may then
use the atomic mass of carbon and Avogadro’s number to ﬁnd the number
of total atoms in the one gram sample. Then will we have the necessary in-
formation to calculate the proportion. First, since we are given the half-life
of
14
6
C, we use eq. 14.13 to obtain the proportionality constant λ:
t
1/2
=
ln 2
λ
⇒λ =
ln 2
t
1/2
λ =
ln 2
5730 yr
=
ln 2
3.012 ×10
9
min
= 2.30 ×10
−10
min
−1
Now to bring this into the rate equation:
15.3 min
−1
= (2.30 ×10
−10
min
−1
)N
0
e
−2.30×10
−10
min
−1
N
0
= 6.65 ×10
10
atoms
14
6
C
Copyright © 2010 by Elsevier, Inc. All rights reserved.
94 14. NUCLEAR PHYSICS - SOLUTIONS
We now calculate the total number of carbon atoms in the sample:
1 g ·
1 mol
12 g
·
6.022 ×10
23
atoms
1 mol
= 5.018 ×10
22
atoms C
We now simply ﬁnd our proportion:
6.65 ×10
10
5.018 ×10
22
=
1.32 ×10
−12
atoms
14
6
C
total atoms C
One check from Appendix B shows that this calculation agrees with the fact
that the isotope of carbon containingj 14 nucleons is not listed as a percent-
age. This indicates its rarity and thus reliability for dating ancient relics
since these nuclear phenomena are readily measurable.
b) We use the value of λ calculated above as well as the value of N
0
which
would have been relevant at the time that the organism was alive. We con-
vert 20,000 years to minutes and use our rate equation with this as the new
time:
20, 000 years = 1.051 ×10
10
minutes
R = (2.30 ×10
−10
min
−1
)(6.65 ×10
10
min) e
−2.30×10
−10
min
−1
(1.051×10
10
min)
= 1.36 beta rays per minute
We see that there are many fewer counts in the ancient sample than in the
current one.
13. We may calculate the release of energy of the
7
4
Be nucleus by ﬁnding the
diﬀerence in binding energy between
7
4
Be and
7
3
He. Since the energy of
7
4
He
is given in table 13.2 to be 39.25 MeV, we need just use the semi-empirical
formula (eq. 14.6) for
7
4
Be:
B(3, 4) = 67a −b7
2/3
−
4
2
d
7
1/3
−s
(−1)
2
7
−0
= 97.899 MeV
We now subtract the two:
∆E = 97.899 −39.25 = 58.65 MeV
Copyright © 2010 by Elsevier, Inc. All rights reserved.
95
Thus the energy released is 58.65 MeV.
15. a) We begin by using equation 14.9 to calculate the nuclear masses
of
9
2
2
34U and
9
0
2
30U:
m(92, 142) = 221710 MeV/c
2
m(90, 140) = 217913 MeV/c
2
Now, using the data from Appendix B, m(2, 2) = 3728.4 MeV/c
2
, and:
∆E = 221710 MeV −217913 MeV −3728.4 MeV = 68.44 MeV
b) The energy carried oﬀ by the alpha-particle will equal the diﬀerence in
binding energies of
9
2
2
34U and
9
0
2
30U. Using equation 14.6,
B(92, 142) = 1923.88 MeV
B(90, 140) = 1883.46 MeV
Thus, the alpha-particle carries oﬀ 40.42 MeV of energy.
17. We follow a very similar procedure to that used in section 14.5. However,
we will use the following subscripts to represent the various particles in the
process: D represents the deuteron, X
∗
represents the excited nucleus, and
p represents the proton. We begin with our statements of conservation of
momentum:
p
D
= p
p
cosθ +P
x
0 = p
p
sinθ +P
y
Solving for P
x
and P
y
:
P
x
= p
D
−p
p
cosθ
P
x
= −p
p
sinθ
The total kinetic energy before the collision is equal to the kinetic energy of
the proton:
E
i
=
p
2
D
2m
D
Copyright © 2010 by Elsevier, Inc. All rights reserved.
96 14. NUCLEAR PHYSICS - SOLUTIONS
The total kinetic energy after the collision is:
E
f
=
p
2
p
2m
p
+
1
m
X
∗
(p
D
−p
p
cosθ)
2
+
1
2m
X
∗
(p
p
sinθ)
2
Due to the conservation of energy, the energy absorbed by the nucleus will
be equal to the loss of kinetic energy:
E = E
i
−E
f
=
p
2
D
2m
D
−
p
2
p
2m
p
−
1
m
X
∗
(p
2
D
+p
2
p
−2p
D
p
p
cosθ)
We now substitute the corresponding energies for the ﬁrst two terms as well
as the following:
p
2
D
2
= E
D
m
D
p
2
p
2
= E
p
m
p
p
D
p
D
2
=
_
E
D
E
p
m
D
m
p
Our expression for the energy of the excited nucleus due to deuteron stripping
is now:
E = E
D
_
1 −
m
D
m
X
∗
_
−E
p
_
1 −
m
p
m
X
∗
_
−
2
_
E
D
E
p
m
D
m
p
m
X
∗
cosθ
Copyright © 2010 by Elsevier, Inc. All rights reserved.

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